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Balancing Redox Equations

For the follo...

Question

For the following reaction scheme, percentage yields are given along the arrow: $x g$ and $y g$ are the mass of $R$ and $U$ , respectively.
(Use : Molar mass (in ${g m o l}^{- 1}$ ) of $H, C$ and $O$ as $1, 12$ and $16$ , respectively)

The value of $x$ is:

A

1.62

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B

1.6

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C

1.620

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Solution

Molar mass of $P = 40$
So moles of $P = \frac{4}{40} = 0.1 mol$
$P (0.1 mol) 75 % - - \to Q ⎛ ⎝ 0.1 \times \frac{3}{4} ⎞ ⎠$

$3 Q ⎛ ⎝ 0.1 \times \frac{3}{4} ⎞ ⎠ 40 % - - \to R ⎛ ⎝ \frac{1}{3} \times 0.1 \times \frac{3}{4} \times 0.4 ⎞ ⎠$

So, moles of $R = 0.01$ mole
Molar mass of $(R) = 162$
So, $x = 0.01 \times 162 = 1.62 g$

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