Question

Antimony can be prepared from its sulphide ore by the following two-step process:
$2S{b}_2{S}_3+9O_2\to S{b}_4{O}_6+6S{O}_2$
$S{b}_4{O}_6+6C\to 4Sb+6CO$
Molar mass of Sb = 121.76 g/mol.
Molar mass of S = 32 g/mol

When 339.52 g of $S{b}_2{S}_3$ is reacted with 112 L of oxygen at STP, 200 g of Sb is obtained. Calculate the percentage yield of the reaction.

• A.
  70%
• B.
  82%
• C.
  95%
• D.
  50%

Answer 1

The correct option is B

82%
Given
Mass of $S{b}_2{S}_3$ = 339.52 g
Moles of $S{b}_2{S}_3$ = $\frac{339.52}{339.52}=1mol$

Moles of oxygen = $\frac{112}{22.4}$ = 5 mol

According to stoichiometry, 2 moles of $S{b}_2{S}_3$ react with 9 moles of oxygen and hence one mole react with 4.5 moles of oxygen. Thus, there will be 0.5 moles of oxygen in excess.

The limiting reagent is $S{b}_2{S}_3$.
Again, from two mole of $S{b}_2{S}_3$ we get four moles of $Sb$.
Thus, from one mole of $S{b}_2{S}_3$ we will get two moles of $Sb$.
Hence, theoretical yield $=2\times 121.76=243.52$
We know that,
Experimental yield = 200 g

$Percentageyield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100=\frac{200}{243.52}\times 100\approx 82\%$