A 1.2 M solution of H2S reacts with a 0.80 M solution of NH2-. What volume of H2S is needed to neutralize 45mL of NH2-?

Answers

Answer 1

The volume of H2S needed to neutralize 45mL of NH2- is 0.053 L

What do you mean by neutralization reaction?

A neutralization reaction, in which an acid and a base react to form water and salt.

The reaction between H2S and NH2- is a neutralization reaction. The balanced equation for the reaction is:

H2S(aq) + NH2-(aq) → NH4HS(aq)

We know that the concentration of H2S is 1.2 M and the concentration of NH2- is 0.80 M and we are trying to find the volume of H2S needed to neutralize 45mL of NH2-.

The number of moles of H2S can be calculated by using the formula:

moles = molarity x volume

The volume of H2S needed to neutralize 45mL of NH2- is equal to the number of moles of NH2- in 45mL, divided by the concentration of H2S.

moles of NH2- = 0.080 L * 0.80 mol/L = 0.064 mol

volume = moles of NH2- / molarity of H2S

volume = 0.064 mol / 1.2 mol/L = 0.053 L

Therefore, the volume of H2S needed to neutralize 45mL of NH2- is 0.053 L.

To know more about neutralization, visit:

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