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Percentage Composition

In the given ...

Question

In the given reaction, $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$
$39 g m$ of $C_{6} H_{6}$ reacts with $40 g m$ of $B r_{2}$ . It was found that $19.625 g m$ of $C_{6} H_{5} B r$ was obtained. Calculate the percentage yield of the reaction.

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 this reaction, $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$
Moles of $C_{6} H_{6}, B r_{2}, C_{6} H_{5} B r$ :
$n_{C_{6} H_{6}} = \frac{39}{78} = 0.5 n_{B r_{2}} = \frac{40}{160} = 0.25 n_{C_{6} H_{5} B r} = \frac{19.625}{157} = 0.125$

We need to find the limiting reagent here, which is the one whose ratio of actual moles and stoichiometric coefficients is the smallest.
$B r_{2}$ is the limiting reagent for this reaction.
From reaction stoichiometry:
Expected $C_{6} H_{5} B r$ formed:
$= 0.25 m o l = 39.25 g m$
$% Y i e l d = \frac{A c t u a l Y i e l d}{T h e o r i t i c a l Y i e l d} \times 100$
$A c t u a l Y i e l d = 19.625 g m$
$T h e o r i t i c a l Y i e l d = 39.25 g m$
$P e r c e n t a g e Y i e l d = \frac{19.625}{39.25} \times 100 = 50 %$

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Percentage Composition

Standard XII Chemistry

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In the given reaction, C6H6 + Br2 → C6H5Br + HBr 39 gm of C6H6 reacts with 40 gm of Br2. It was found that 19.625 gm of C6H5Br was obtained. Calculate the percentage yield of the reaction.

In the given reaction, $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$
$39 g m$ of $C_{6} H_{6}$ reacts with $40 g m$ of $B r_{2}$ . It was found that $19.625 g m$ of $C_{6} H_{5} B r$ was obtained. Calculate the percentage yield of the reaction.