Question

(a) If 112 cm³ of hydrogen sulphide is mixed with 120 cm³ of chlorine at STP, what is the mass of sulphur formed?
H₂S + Cl₂ $\stackrel{}{\to }$ 2HCl + S
(b) Washing soda has the formula Na₂CO₃·10H₂O. What is the mass of anhydrous sodium carbonate left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
(c) When excess lead nitrate solution was added to a solution of sodium sulphate, 15.1 g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
Na₂SO₄ + Pb(NO₃)₂ $\stackrel{}{\to }$ PbSO₄ + 2NaNO₃
(H = 1; C = 12; O = 16; Na = 23; S = 32; Pb = 207)

Answer 1

(a) Moles of hydrogen sulphide in 112 cm³ at STP = $\frac{\mathrm{Volume}\mathrm{of}{\mathrm{H}}_2\mathrm{S}}{\mathrm{Molar}\mathrm{Volume}}=\frac{112}{22400}\mathrm{mol}=0.0050$
Moles of chlorine in 120 cm³ at STP = $\frac{\mathrm{Volume}\mathrm{of}{\mathrm{Cl}}_2}{\mathrm{Molar}\mathrm{Volume}}=\frac{120}{22400}\mathrm{mol}=0.0053$

Hence, hydrogen sulphide is the limiting reagent. Mass of sulphur formed will depend upon the amount of hydrogen sulphide available for the reaction.
According to the balanced chemical equation, one mole of hydrogen sulphide will produce one mole of sulphur.
Thus, 0.005 moles of hydrogen sulphide will produce 0.005 moles of sulphur.
Mass of sulphur formed = Moles of sulphur $\times$ Atomic mass of sulphur
= 0.005 $\times$ 32
= 0.16 g

(b) Molecular mass of Na₂CO₃·10H₂O = 2(23) + 12 + 3(16) + 20(1) + 10(18) = 286 g
Number of moles in 57.2 g of Na₂CO₃·10H₂O = $\frac{57.2}{286}=0.2$
Mass of water of crystallisation in 0.2 mol of washing soda = 0.2 $\times$ 10(Molecular mass of water)
= 0.2 $\times$ 10(18)
= 36 g
Mass of anhydrous sodium carbonate left when all the water of crystallisation is expelled = 57.2 − 36 = 21.2 g

(c) First, let us determine the limiting reagent in the given reaction. As lead nitrate is present in excess; therefore, sodium sulphate is the limiting reagent.
According to the balanced chemical equation, number of moles of lead sulphate produced is equal to number of moles of sodium sulphate used.
Number of moles of lead sulphate formed = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{lead}\mathrm{sulphate}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{lead}\mathrm{sulphate}}=\frac{15.1}{303}=0.05$ mol
Number of moles of sodium sulphate present in original solution = 0.05 mol
Mass of sodium sulphate present in original solution = Number of moles $\times$ Molecular mass of sodium sulphate
= (0.05 $\times$ 142) g
= 7.1 g