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Stoichiometric Calculations

S + O - 2. `r...

Question

$S + O_{2} \to S O_{2} (Yield = 80%)$
$S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)$
$H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)$
8 g of Sulphur is burnt to form $S O_{2}$ , which is oxidized by $C l_{2}$ water. The solution is then treated with $B a C l_{2}$ solution. The amount of $B a S O_{4}$ precipitated is:
(Molar mass of Barium = $137 g/mol$ )

8.39 g

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18.39 g

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24.26 g

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13.67 g

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Solution

The correct option is A $8.39 g$
$S + O_{2} \to S O_{2}$ $(I) (Yield = 80%)$
$S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l$ $(I I) (Yield = 60%)$
$H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l$ $(I I I) (Yield = 30%)$
In reaction (i),
1 mol of $S$ forms 1 mol of $S O_{2}$
Again, $8 g$ sulphur = $\frac{8}{32}$ mol of $S$ = 0.25 mol $S$
Yield of the reaction is given as 80 %
Actual moles of $S O_{2}$ formed $= 0.25 \times 0.80 = 0.2 m o l$
In reaction (ii),
1 mol of $S O_{2}$ forms 1 mol of $H_{2} S O_{4}$
0.2 mol of $S O_{2}$ will form 0.2 mol of $H_{2} S O_{4}$
Yield of the reaction is given as 60 %
Actual amount of $H_{2} S O_{4}$ formed $= 0.2 \times 0.60 = 0.12 m o l$
In reaction (iii),
1 mol of $H_{2} S O_{4}$ reacts to form 1 mol of $B a S O_{4}$
0.12 mol of $H_{2} S O_{4}$ reacts to form 0.12 mol of $B a S O_{4}$
Yield of the reaction is given as 30 %
So moles of $B a S O_{4}$ produced = $0.12 \times 0.3 = 0.036 m o l$
Molar mass of $B a S O_{4} = 137 + 32 + 64 = 233 g$
Actual amount of precipitate obtained $0.036 \times 233 g = 8.39 g$

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S + O_{2} \to S O_{2} (Yield = 80%)

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H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)

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