# Iron weighing 14 g was fused with sulfur weighing 4.8 g, and hcl was added to the resulting mixture.

**Iron weighing 14 g was fused with sulfur weighing 4.8 g, and hcl was added to the resulting mixture. What gases are released and what are their volumes?**

We write down the equation of reactions.

Fe + S = FeS.

We find the amount of the substance iron and sulfur. We use the following formula.

n = m / M.

We calculate the molar mass of iron and sulfur.

M (Fe) = 56 g / mol.

M (S) = 32 g / mol.

We find the amount of substance.

n (Fe) = 14 g / 56 g / mol = 0.25 mol.

n (S) = 4.8 g / 32 g / mol = 0.15 mol.

This means that iron is in excess, which means we will count by sulfur.

We write down the uranium of reactions.

0.15 mol S = 0.15 mol FeS.

FeS + 2 HCL = FeCl 2 + H2S.

Next, we find the amount of hydrogen sulfide substance by the reaction equation.

0.15 mol FeS – х mol H2S

1 mol FeS – 1 mol H2S

Hence, n (H2S) = 0.15 mol.

Next, we find the volume.

n = V / Vm.

V = n × Vm = 0.15 mol × 22.4 L / mol = 3.36 L.

Answer: hydrogen sulfide H2S gas with a volume of 3.36 liters will be released.