What volume of hydrogen sulfide will be released when 0.88 g of iron (II) sulfide is treated with an excess of hydrochloric acid?

1.Let’s find the amount of iron (II) sulfide – FeS.

M (FeS) = 56 + 32 = 88 g / mol.

n = m: M

n = 0.88 g: 88 g / mol = 0.01 mol.

2. Let’s compose the equation of the reaction between iron sulfide and hydrochloric acid.

FeS + 2HCl = FeCl2 + H2S.

According to the reaction equation, for 1 mole of iron sulfide, there is 1 mole of gas – hydrogen sulfide, that is, their quantitative ratios are 1: 1. Then the amount of iron sulfide substance will be the same with the amount of hydrogen sulfide substance.

n (H2S) = 0.01 mol.

Let’s find the volume by the formula:

V = n Vn, where Vn is the molar volume of the gas, equal to 22.4 l / mol.

V = 22.4 L / mol x 0.01 mol = 0.224 L.

Answer: V = 0.224 l.