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Q. In the following reaction, $P$ gives two products $Q$ and $R$ each in $40\%$ yield.
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If the reaction is carried out with $420 \,mg$ of $P$ the reaction yields $108.8\,mg$ of $Q$. The amount of $R$ produced in the reaction is closest to

KVPYKVPY 2019

Solution:

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Number of moles of P $=\frac{420\times10^{-3}}{210} $
$=2\times10^{-3}$
Number of moles of $Q$ and $R = 40 \%$ of
$P=\frac{40}{100}\times2\times10^{-3} $
$=8\times10^{-4}$
mass of
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$=8\times10^{-4}\times10^{6}=84.8\times10^{-3}$
$=84.8\,mg$
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$=8\times10^{-4}\times136$
$=108.8\times10^{-3}$
$=108.8\,mg$
which is $Q$
$\therefore $ Mass of $R=84.8\, mg$