Question

X g of a pure sample of iron (II) sulphide reacted completely with excess dilute hydrochloric acid to give 3.2 g of iron (III) chloride according to the equation blow . FeS + 2HCl = FeCl2 + H2S

Answer 1

2FeS + 6HCl → 2FeCl₃ + 2H₂S + H₂

M(FeCl₃) = 162.2 g/mol

n(FeCl₃) $= \frac{m}{M} = \frac{3.2}{162.2} = 0.01979 \;mol$

According to the reaction equation:

n(FeS) = n(FeCl₃) = 0.01979 mol

M(FeS) = 87.91 g/mol

m(FeS) $= 87.91 \times 0.01979 = 1.73 \;g$

Answer: X = 1.73 g