Answer to Question #304779 in Chemistry for Blake

Question

What volume of hydrogen sulfide gas is formed at 0°C and 100 kPa when 1.24g iron(II) sulfide is reacted with 30 mL of 1.04 mol L^-1 hydrochloric acid solution? Iron(II) chloride is the other product.

Answer is 0.32 L (2 sf)

Answer 1

Solution:

iron(II) sulfide = FeS

hydrochloric acid = HCl

hydrogen sulfide = H₂S

iron(II) chloride = FeCl₂

Calculate the moles of each reactant:

1) The molar mass of FeS is 87.91 g/mol

Therefore,

Moles of FeS = (1.24 g FeS) × (1 mol FeS / 87.91 g FeS) = 0.0141 mol FeS

Moles of FeS = 0.0141 mol

2) Molarity of HCl = Moles of HCl / Liters of solution

Therefore,

Moles of HCl = Molarity of HCl × Liters of solution = (1.04 mol L⁻¹) × (0.03 L) = 0.0312 mol HCl

Moles of HCl = 0.0312 mol

Balanced chemical equation:

FeS(s) + 2HCl(aq) → FeCl₂(s) + H₂S(g)

According to stoichiometry:

1 mol of FeS reacts with 2 mol of HCl

Thus, 0.0141 mol of FeS reacts with:

(0.0141 mol FeS) × (2 mol HCl / 1 mol FeS) = 0.0282 mol HCl

However, initially there is 0.0312 mol of hydrochloric acid (HCl) (according to the task).

Thus, FeS acts as limiting reactant and HCl is excess reactant.

According to stoichiometry:

1 mol of FeS produces 1 mol of H₂S

Thus, 0.0141 mol of FeS produces:

(0.0141 mol FeS) × (1 mol H₂S / 1 mol FeS) = 0.0141 mol H₂S

Moles of H₂S = 0.0141 mol

Ideal gas equation can be used.

PV = nRT

P = 100 kPa

V = unknown

n(H₂S) = 0.0141 mol

R = 8.314 L kPa K⁻¹ mol

T = 0°C + 273.15 = 273.15 K

Thus,

V = nRT / P

V(H₂S) = (0.0141 mol × 8.314 L kPa K⁻¹ mol × 273.15 K) / (100 kPa) = 0.32 L

V(H₂S) = 0.32 L

Answer: 0.32 liters of hydrogen sulfide gas (H₂S) is formed