Chemistry College

Calculate how many grams of rust (Fe2O3) should form when 10.0g of iron reacts with 20.0g of oxygen.
4Fe + 3O2 → 2Fe2O3

Answers

Answer 1

Taking into account the reaction stoichiometry, 14.31 grams of Fe₂O₃ are formed when 10.0g of iron reacts with 20.0g of oxygen.

Reaction stoichiometry

In first place, the balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe: 4 molesO₂: 3 molesFe₂O₃: 2 moles

The molar mass of the compounds is:

Fe: 55.85 g/moleO₂: 32 g/moleFe₂O₃: 159.7 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe: 4 moles ×55.85 g/mole= 223.4 gramsO₂: 3 moles ×32 g/mole= 96 gramsFe₂O₃: 2 moles ×159.7 g/mole= 319.7 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 96 grams of O₂ reacts with 223.4 grams of Fe, 20 grams of O₂ reacts with how much mass of Fe?

mass of Fe= (20 grams of O₂ ×223.4 grams of Fe) ÷96 grams of O₂

mass of Fe= 46.54 grams

But 46.54 grams of Fe are not available, 10 grams are available. Since you have less mass than you need to react with 20 grams of O₂, Fe will be the limiting reagent.

Mass of Fe₂O₃ formed

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 223.4 grams of Fe form 319.7 grams of Fe₂O₃, 10 grams of Fe form how much mass of Fe₂O₃?

mass of Fe₂O₃= (10 grams of Fe×319.7 grams of Fe₂O₃)÷223.4 grams of Fe

mass of Fe₂O₃= 14.31 grams

Finally, 14.31 grams of Fe₂O₃ are formed.

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Related Questions

A gas at a pressure of 2.0 atm occupies a volume of 20 liters. If the volume is decreased to 10 liters and the pressure is increased to 4.0 atm, what will be the final volume of the gas?

Answers

The initial condition of the gas can be expressed using the combined gas law:

P1V1 / T1 = P2V2 / T2

where P1 = 2.0 atm, V1 = 20 L, and T1 is the initial temperature (which we will assume remains constant).

If the volume is decreased to 10 L and the pressure is increased to 4.0 atm, the new conditions can be expressed as:

P2 = 4.0 atm, V2 = ?, and T2 = T1

We can rearrange the combined gas law to solve for V2:

V2 = (P1V1T2) / (P2T1)

Plugging in the given values, we get:

V2 = (2.0 atm x 20 L x T1) / (4.0 atm x T1) = 10 L

Therefore, the final volume of the gas is 10 liters.

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Which of the these is a balanced chemical equation?

Question 18 options:

H2O + CHO2 → H2CO3


H2O + CO2 → H2CO3


3H2O + 2CO2 → 2H2CO3


2H2O + 2CO2 → H4CO4

Answers

Answer:

The balanced chemical equation is:

3H2O + 2CO2 → 2H2CO3

Explanation:

For electrical currents, the switch must be in the
a. Closed
b. Open
c. Reverse
d. Doesn't matter
position for the current to flow. (3.3.1)

Answers

Closed because it will flow through

26 What is the charge on each ion in these compounds?
(a) CaS
(b) MgF2
(c) Cs,O
(d) ScCl,
(e) Al,S,

Answers

The charges present on the following ionic compounds are Ca²⁺, Mg²⁺, Cs⁺, Sc⁺, Al³⁺, S²⁻, F⁻, O²⁻, Cl⁻.

Ionic compounds are held together by ionic bonds are classed as ionic compounds. Elements can gain or lose electrons in order to attain their nearest noble gas configuration. The formation of ions (either by gaining or losing electrons) for the completion of octet helps them gain stability.

In a reaction between metals and non-metals, metals generally loose electrons to complete their octet while non-metals gain electrons to complete their octet. Metals and non-metals generally react to form ionic compounds.

Ionic compounds include salts, oxides, hydroxides, sulphides, and the majority of inorganic compounds. Ionic solids are held together by the electrostatic attraction between the positive and negative ions.

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Given:
180.0 mL chloric acid (HCIO3)
440.0 mL of 1.75 M strontium hydroxide (Sr(OH)2)
Wanted: [HCIO3] necessary to neutralize Sr(OH)?

Answers

The molarity of the 180.0 mL chloric acid, HClO₃ solution needed to neutralize the 440.0 mL of 1.75 M strontium hydroxide, Sr(OH)₂ is 8.56 M

How do i determine the molarity of the chloric acid, HClO₃?

We'll begin by writing the balanced equation for the reaction. This is given below:

2HClO₃ + Sr(OH)₂ —> Sr(ClO₃)₂ + 2H₂O

The mole ratio of the acid, HClO₃ (nA) = 2The mole ratio of the base, Sr(OH)₂ (nB) = 1Volume of Sr(OH)₂ (Vb) = 440.0 mLMolarity of Sr(OH)₂ (Mb) = 1.75 M Volume of HClO₃ (Va) = 180.0 mLMolarity of HClO₃ (Ma) =?

The molarity of the chloric acid, HClO₃ solution necessary can be obtained as follow:

MaVa / MbVb = nA / nB

(Ma × 180) / (1.75 × 440) = 2

Cross multiply

Ma × 180  = 1.75 × 440 × 2

Divide both side by 180

Ma = (1.75 × 440 × 2) / 180

Ma = 8.56 M

Thus, we can conclude that the molarity of the chloric acid, HClO₃ solution is 8.56 M  

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Complete question:

Given that 180.0 mL chloric acid (HCIO3) reacted with 440.0 mL of 1.75 M strontium hydroxide (Sr(OH)2). What is the molarity of HCIO3 necessary to neutralize Sr(OH)?

Wanted: [HCIO3] necessary to neutralize Sr(OH)?

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