Question

Iron (III) sulfide reacts with gaseous hydrogen chloride to produce iron (III) chloride and hydrogen sulfide. How much iron (III) sulfide would be required to react with 1.5 mol HCl?

Options:
a) 0.75 mol
b) 1.5 mol
c) 2.0 mol
d) 3.0 mol

Answer 1

The answer to how much iron (III) sulfide is required to react with 1.5 mol hydrogen chloride would be 0.25 moles, which is not included in the provided options. This result is based on the assumed stoichiometric ratios from a typical reaction between a metal sulfide and hydrogen chloride.

Step-by-step explanation:

To find out how much iron (III) sulfide is required to react with 1.5 mol of hydrogen chloride (HCl), we need a balanced chemical equation. Unfortunately, we do not have the balanced equation for the reaction between iron (III) sulfide and hydrogen chloride given in the information. However, based on the common stoichiometry in reactions between a metal sulfide and hydrogen chloride, we can assume that the reaction proceeds as follows:

Fe2S3 + 6HCl → 2FeCl3 + 3H2S

To find how much iron (III) sulfide would react with 1.5 mol of HCl, we would look at the mole ratios provided in the balanced equation. From the equation, we see that 6 moles of HCl react with 1 mole of iron (III) sulfide. Since questions pertain to reactions where 1.5 moles of HCl are used, we would set up the stoichiometry as follows:

(Moles of Fe2S3) = (Moles of HCl) ÷ (Mole ratio of HCl to Fe2S3)

(Moles of Fe2S3) = 1.5 moles HCl ÷ 6 moles HCl/mole Fe2S3

When we perform the calculation, we get:

(Moles of Fe2S3) = 0.25 moles Fe2S3

Therefore, the correct answer is not included in the options provided.