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Percentage Composition

In the given ...

Question

In the given reaction, $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$
$39 g$ of $C_{6} H_{6}$ reacts with $40 g$ of $B r_{2}$ . It was found that $19.625 g$ of $C_{6} H_{5} B r$ was obtained. Calculate the percentage yield of the reaction.
$(Given, Molar mass of: C_{6} H_{6} = 78 {g mol}^{- 1}, B r_{2} = 160 {g mol}^{- 1}, C_{6} H_{5} B r = 157 {g mol}^{- 1})$

50 %

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85 %

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42.5 %

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33.3 %

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Solution

The correct option is A $50 %$
In the reaction, $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$
Moles of $C_{6} H_{6}, B r_{2}, C_{6} H_{5} B r$ :
$n_{C_{6} H_{6}} = \frac{39}{78} = 0.5 n_{B r_{2}} = \frac{40}{160} = 0.25$
We need to find the limiting reagent. It is the one with the smallest ratio of actual moles and stoichiometric coefficients.
To find the limiting reagent,
For $C_{6} H_{6}$ ,
$\frac{number of moles of C_{6} H_{6}}{Stiochiometric coefficient} = \frac{0.5}{1} = 0.5$
For $B r_{2}$
$\frac{number of moles of B r_{2}}{Stiochiometric coefficient} = \frac{0.25}{1} = 0.25$
Hence,
$B r_{2}$ is the limiting reagent for this reaction.
From reaction stoichiometry:
$Expected C_{6} H_{5} B r formed = 0.25 mol = 39.25 g$
$% Yield = \frac{Actual Yield}{Theoretical Yield} \times 100$
$Actual Yield = 19.625 g$
$Theoretical Yield = 39.25 g$
$Percentage Yield = \frac{19.625}{39.25} \times 100 = 50 %$

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