An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.
The kind of indicator used depends on the nature of acid/base reacted.
In the case of CH3NH2 with HBr which strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl orange, methyl red, and chlorophenol blue.
In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.
In the case of HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and phenolphthalein.
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Select True or False: The equilibrium constant for the chemical equation 2NO(g) O2(g) 2NO2(g) is two times the equilibrium constant for the chemical equation NO(g) 1/2O2(g) NO2(g).
Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
The standard redox potentials of isolated components of an electron transport chain in a cyanobacterium are found to be as follows:
Complex A: standard redox potential: -100 mV
Complex B: standard redox potential: -780 mV
Complex C: standard redox potential: +510 mV
Complex D: standard redox potential: +310 mV
Plastocyanin: standard redox potential: +360 mV
Which complex will likely have a binding site with high affinity for reduced plastocyanin?
A. Complex A.
B. Complex B.
C. Complex C.
D. Complex D.
Answer:
B. Complex B.
Explanation:
Complex B will have binding site with high affinity for reduced plastocyanin due to greater redox potential. The high number of redox potential will will transport electron chain in cyanobacteria.
Redox potential is the measure of the electron gain or loss to the electrode. For reduced plastocyanin complex B will have the highest affinity.
What is electron affinity?Electron affinity is the energy released when the atom gets attached to the atom or other molecule. The high number of redox potential increases the electron transport in the cell.
The greater the redox potential more will be its tendency to show electron affinity. To bond with reduced species, the oxidized species must have greater redox potential.
Therefore, option B. complex b will have the highest affinity.
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Match each land resource to its use.
clay - used to make steel
iron ore - used to make batteries
salt - used to make pottery and tiles
aggregate - used in construction
graphite - used as a flavoring in food
i will give 10 points and brainliest!!!
Answer:
answer in picture
Explanation:
90
1
39
is the
In the following decay equation,
90
Sy →
38
et
90
39
-1
A. alpha particle
B. parent element
C. daughter element
D. beta particle
Answer:
D. beta particle
Explanation:
Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.
How many grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride?
Answer:
it is 11.55 and ik because I just had that question
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
Let's consider the following balanced equation.
4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)
The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 0.125 moles of C₈H₄N₂ are:
[tex]0.125 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.0313 molCu(C_{32}H_{16}N_8)[/tex]
The molar mass of Cu(C₃₂H₁₆N₈) is 576.07 g/mol. The mass corresponding to 0.0313 moles of Cu(C₃₂H₁₆N₈) is:
[tex]0.0313 moles \times \frac{576.07g}{mol} = 18.0 g[/tex]
18.0 grams of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 0.125 moles of phthalonitrile in the presence of excess copper(II) chloride.
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How are all compounds similar?
A. They are all made up of ions that are held together by attractions.
B. They are all made up of the same few elements.
C. They are all made up of atoms of two or more different elements.
D. They are all made up of atoms that share electrons.
Answer:
the answer is C
Explanation:
a molecule can be made up of two atoms of the same kind, as when two oxygen atoms bind together to make an oxygen molecule
I'd really appreciate a brainleast
What is Avogadro's number?
O A. 6.02 x 10-23
O B. 6.0223
C. 6.02 x 10
D. 6.02 x 1023
Answer:
E.6.02 10-23Explanation:
Answer:
6.02×10^23 I hope it helps you
Explanation:
I hope it helps you
If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?
After the double replacement reaction from mixing 50.0g of sulfuric acid and 40.0 grams of barium chloride is complete, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain.
The equation of the reaction between sulfuric acid and barium chloride is
BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
From this equation of reaction, it means 1 mole of barium chloride will completely react with 1 mole of sulfuric acid.
From the question, we have 50.0g of sulfuric acid and 40.0 grams of barium chloride.
First, we will determine the number of moles of the sulfuric acid and barium chloride present.
Number of moles is given by the formula
Number of moles = Mass / Molar mass
For sulfuric acid
Mass = 50.0 g
Molar mass = 98.079 g/mol
∴ Number of moles = 50.0 / 98.079
Numbers of moles of sulfuric = 0.509793 mol
For barium chloride
Mass = 40.0 grams
Molar mass of barium chloride = 208.23 g/mol
∴ Number of moles = 40.0 / 208.23
Number of moles of barium chloride = 0.192095 mol
Since the number of moles of sulfuric acid is more than that of barium chloride, then the limiting reagent is barium chloride and the excess reagent is sulfuric acid
NOTE: A limiting reagent is the reactant that is completely used up in a reaction, and it determines when the reaction stops.
Hence, barium chloride will be used up during the reaction (that is, 0 grams will remain after the reaction is complete).
For the mass of sulfuric acid that will remain,
First, we will determine the number of mole that will remain.
Since 1 mole of barium chloride completely reacts with 1 mole of sulfuric acid, then 0.192095 mol of barium chloride will react with 0.192095 mol of sulfuric acid.
∴ The remaining number moles of sulfuric acid = 0.509793 mol - 0.192095 mol
The remaining number moles of sulfuric acid = 0.317698 mol
Then, from
Mass = Number of moles × Molar mass
Mass = 0.317698 mol ×98.079 g/mol
Mass = 31. 1595 gram
Mass ≅ 31.16 grams
∴ 31.16 grams remains after the reaction is complete.
Hence, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain after the double replacement reaction is complete.
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What do scientist use to form a hypothesis
Answer:
an if/then statement
Explanation:
Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
The mass of an empty flask plus stopper is 64.232g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557. Next, water is added to the flask containing the metal and the mass is found to be 226.196. What is the density of the metal
Answer:
11.76 g/cm^3
Explanation:
Mass of empty flask and stopper = 64.232g
Mass of flask filled with water = 153.617 g
Mass of water = 153.617 g - 64.232g = 89.385 g
Mass of flask, stopper and metal = 143.557 g
Mass of metal = 143.557 g - 64.232g = 79.325 g
Mass of water, flask, stopper and metal = 226.196 g
Mass of water = 226.196 g - 143.557 g = 82.639 g
Since mass of water =volume of water
Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3
Density of metal = mass/volume = 79.325 g/6.746 cm^3
= 11.76 g/cm^3
A sample of aluminum, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at .Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to significant digits.
Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
[tex]M=58g[/tex]
Explanation:
From the question we are told that:
Heat Capacity [tex]H=0.897[/tex]
Mass of water [tex]M=200g[/tex]
Initial Temperature of Aluminium [tex]T_a=85.6[/tex]
Initial Temperature of Water [tex]T_{w1}=16.0[/tex]
Final Temperature of Water [tex]T_{w2}=16.0[/tex]
Generally
Heat loss=Heat Gain
Therefore
[tex]M*0.897*(85.6-20.1) =200*4.184*(20.1-16)[/tex]
[tex]M=58g[/tex]
oxidation number of Ni in Ni(CO)4 is
Answer:
0
Explanation:
answer from gauth math
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge.
b. A beta particle contains neutrons.
c. A beta particle is less massive than a gamma ray.
d. A beta particle is a high-energy electron.
Answer:
a. A beta particle has a negative charge. d. A beta particle is a high-energy electron.
Explanation:
Identify the correct descriptions of beta particles.
a. A beta particle has a negative charge. YES. A beta particle is originated in the following nuclear reaction: ¹₀n ⇒ ¹₁H + ⁰₋₁e (beta particle.)
b. A beta particle contains neutrons. NO. It is a electron originated in the nucleus.
c. A beta particle is less massive than a gamma ray. NO. Gamma rays don't have mass while a beta particle has a mass which is half of one thousandth of the mass of a proton.
d. A beta particle is a high-energy electron. YES. Beta particles are nuclear originated hig-energy electrons.
The idea that the behavior of the states of matter is determined by the kinetic energy and movement of their particles is called _____…
A. Sublimation Theory
B. Kinetic Movement Theory
C. Kinetic Molecular Theory
D. Van der Waals Theory
Answer:
C . Kinetic Molecular Theory
What is the difference between elimination and substitution reaction
Identify the key factors that will determine if a reaction undergoes elimination or substitution mechanism.
Use the following reagents to determine the type of reaction pathway expected and determine the products in each reaction.
a. Tert BuO- in tertbutanol and chlorobutane
b. KOH in water and bromobutane
c. NaI in acetone and bromobutane
Write a conclusion of no more than two paragraphs to summarize your results
Answer:
a) E2
b) SN2
c) SN2
Explanation:
A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.
We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.
Complete and balance the nuclear equations for the following fission reactions.
a. 23592U+10n→16062Sm+7230Zn+?10n. Express your answer as a nuclear equation.
b. 23994Pu+10n→14458Ce+?+210n
Answer:
23592U+10n→14454Xe+9038Sr+210n
Explanation:
a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.
It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much concentrated nitric acid is required to make the desired solution?
Explanation:
The required concentration of [tex]HNO_3[/tex] M1 =0.222 M.
The required volume of [tex]HNO_3[/tex] is V1 =225 mL.
The standard solution of [tex]HNO_3[/tex] is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles [tex]n=molarity * volume[/tex]
[tex]M_1.V_1=M_2.V_2[/tex]
[tex]V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL[/tex]
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
How many protons does Tin have?
A. 50
B. 68
C. 118
Hello There!
Tin has 50 protons.Hope that helps you!
~Just a felicitous girlie
#HaveASplendidDay
[tex]SilentNature[/tex]
The reaction for photosynthesis producing glucose sugar and oxygen gas is:
__CO2(g) + __H2O(l) UV/chlorophyl−→−−−−−−−−−−−−−− __C6H12O6(s) + __O2(g)
What is the mass of glucose (180.18 g/mol) produced from 2.20 g of CO2 (44.01 g/mol)?
a. 66.1 g C6H12O6
b. 396 g C6H12O6
c. 54.0 g C6H12O6
d. 1.50 g C6H12O6
e. 9.01 g C6H12O6
The correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.
To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
So, in this case, the balanced reaction is:
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:
CO₂: 6 moles H₂O: 6 moles C₆H₁₂O₆: 1 mole O₂: 6 molesSo, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:
[tex]moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}[/tex]
moles of CO₂= 0.05 moles
Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?
[tex]moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}[/tex]
moles of C₆H₁₂O₆= 8.33*10⁻³
Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:
[tex]mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}[/tex]
mass of glucose= 1.5 grams
In summary, the correct option is d.: 1.5 grams of glucose is produced from 2.20 g of CO₂.
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313.9 liters of a gas has a pressure of 390.89 kPa at 76.6°C. If the pressure increases to 718.3 kPa and the temperature to 154.2°C, what would be the new volume of the gas?
A.) 210
B.) 353
C.) 470
D.) 209.92
Explanation:
P1V1/T1 = P2V2/T2
=((390.89×313.9)/76.6)×((718.3×V2)/154.2)
= (122700.371/76.7) × ((718.3×V2)/154.2)
make V2 the subject of the formula...
V2 =(122700.371×154.2)/(76.6×718.3)
V2 =18920397.21/55021.78
V2 = 343.87
Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water
Answer:
% NiCl2 = 24.13%
% water = 78.57%
Explanation:
Mass percentage = mass of solute/mass of solution × 100
According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:
Mass of solution = 159g + 500g
Mass of solution = 659g
Therefore;
A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100
% Mass of NiCl2 in solution = 159/659 × 100
% Mass of NiCl2 in solution = 0.2413 × 100
= 24.13%
B) % Mass of water in solution = mass of water/mass of solution × 100
% Mass of water in solution = 500/659 × 100
% Mass of water in solution = 0.7587 × 100
% Mass of water in solution = 75.87%
The following reaction is not an oxidation-reduction reaction: Fe(s) + 2Hl(aq) --- Fel (aq) + H_(8) Select one: O True O False
Explanation:
the reaction is indeed an oxidation reduction reaction
Name the following compound: CH3CH2CH2CH2CH2CH2CCCH3
2-nonene
7-nonyne
2-heptyne
2-nonyne
Finding Nemo??? sorry I really need these points
Answer:
2-nonyne
explanation:
First consider the type of bonds, there are tripple bonds of carbon to carbon, in position 2 from the right.
hence it us alkyn.
There are 9 carbons.
Do you think that the human being is the center of the universe?
Net ionic reaction of H2SO4 with Ba(OH)2
Answer:
This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.
Calculate the vapor pressure of a solution made by dissolving 550 grams of glucose (molar mass = 180.2 g/mol) in 1020.0 ml of water at 25°C. The vapor pressure of pure water at 25°C is 23.76 mm Hg. Assume the density of the solution is 1.00 g/ml. (760 torr = 760 mmHg = 1 atm)
Answer:
22.55 mmHg (0.03 atm)
Explanation:
According to Raoult's law, the vapor pressure (Psolution) of a solution is given by:
Psolution = Xsolvent x Psolvent
Where Xsolvent is the mole fraction of the solvent in the solution and Psolvent is the vapor pressure of the pure solvent.
From the data, we have: Psolvent = 25.76 mmHg
We have to calculate Xsolvent, which is equal to the moles of solvent divided into the total number of moles.
The solution is composed of the solute (glucose) dissolved in the solvent (water). So, the total number of moles is calculated from the moles of solute and solvent.
To calculate the moles of solute (glucose), we divide the mass of glucose into its molar mass:
moles of glucose = mass/molar mass = 550 g/(180.2 g/mol) = 3.05 mol
The same for the moles of solvent (water). The mass of water is obtained from the product of the volume and density:
mass of water = volume x density = 1020.0 mL x 1.00 g/mL = 1020.0 g
molar mass H₂O = (1 g/mol x 2) + 16 g/mol = 18 g/mol
moles of water = mass water/molar masss = 1020.0 g/(18 g/mol) = 56.67 mol
Now, we can calculate Xsolvent:
Xsolvent = moles of water/total moles
total moles = moles glucose + moles water = 3.05 mol + 56.67 mol = 59.72 mol
⇒ Xsolvent = 56.67 mol/(59.72 mol) = 0.9489
Finally, we calculate the vapor pressure of the solution:
Psolution = 0.9489 x 23.76 mmHg = 22.55 mmHg
22.55 mmHg x 1 atm/760 mmHg = 0.03 atm
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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Hydrogen is manufactured on an industrial scale by this sequence of reactions:
CH4(g) + H2O(g) ⇆ CO(g) + 3H2(g)
CO(g)+ H2O(g) ⇆ CO(g) + H2(g)
The net reaction is:
CH4(g) + H2O(g) ⇆ CO(g) + 4H2(g)
Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K1 and K2.
Answer:
[tex]K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
[tex]K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}[/tex]
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
[tex]K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}[/tex]
Regards!
For the following reaction, 4.77 grams of carbon (graphite) are allowed to react with 16.4 grams of oxygen gas.
carbon (graphite) (s) + oxygen (g) → carbon dioxide (g)
1. What is the maximum amount of carbon dioxide that can be formed?
2. What is the FORMULA for the limiting reagent?
3. What mass of the excess reagent remains after the reaction is complete?
Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
