Xg of a pure sample of iron (II) sulphide reacted completely with excess dilute hydrochloric acid to give 3.20g of iron (II) chloride according to the follo...
Question 1 Report
Xg of a pure sample of iron (II) sulphide reacted completely with excess dilute hydrochloric acid to give 3.20g of iron (II) chloride according to the following equation: FeS\(_{(s)}\) + 2HCI\(_{(aq)}\) ---> FeCl\(_{2(aq)}\) + H\(_2\)S\(_{(g)}\).
(a) Mention one method apart from heating by which the reaction can be made to proceed faster
(b) Calculate the value of X. [CI = 35.5, Fe = 56; FeS = 88g mol\(^{-1}\)]
Answer Details
(a) One method (apart from heating) by which the reaction can be made faster: increase the surface area of the iron(II) sulphide by using it in a finely powdered form (using a more concentrated acid is also acceptable).
(b) Calculating X. Equation: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow FeCl_{2(aq)} + H_2S_{(g)}\]
Molar mass of \(FeCl_2 = 56 + (2 \times 35.5) = 56 + 71 = 127\,g\,mol^{-1}\)
\[\text{moles of } FeCl_2 = \frac{3.20}{127} = 0.0252\,mol\]
From the equation, 1 mole \(FeS\) gives 1 mole \(FeCl_2\), so moles of \(FeS = 0.0252\,mol\).
\[X = \text{mass of } FeS = 0.0252 \times 88 = 2.22\,g\]
So X = 2.22 g (approximately).
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