In the following reaction,P gives two product | vedclass

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(C) The reaction is an ozonolysis of an alkene $P$ $(M.wt. = 210)$.Number of moles of $P = \frac{420 	imes 10^{-3} \,g}{210 \,g/mol} = 2 	imes 10^{-

Vedclass · Accepted Answer

$84.8$

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(C) The reaction is an ozonolysis of an alkene $P$ $(M.wt. = 210)$.Number of moles of $P = \frac{420 	imes 10^{-3} \,g}{210 \,g/mol} = 2 	imes 10^{-3} \,mol$.Since the yield of both $Q$ and $R$ is $40 \,\%$,the number of moles of $Q$ and $R$ produced is $40 \,\%$ of $2 	imes 10^{-3} \,mol = 8 	imes 10^{-4} \,mol$.Product $Q$ is benzaldehyde ($C_6H_5CHO$,$M = 106$) and $R$ is $3$-methoxybenzaldehyde ($CH_3OC_6H_4CHO$,$M = 136$).Mass of $R = 	ext{moles} 	imes 	ext{molar mass} = 8 	imes 10^{-4} \,mol 	imes 136 \,g/mol = 1088 	imes 10^{-4} \,g = 108.8 \,mg$.Mass of $Q = 8 	imes 10^{-4} \,mol 	imes 106 \,g/mol = 848 	imes 10^{-4} \,g = 84.8 \,mg$.Given that $108.8 \,mg$ of $Q$ is produced,$R$ must be the product with mass $84.8 \,mg$.

In the following reaction,$P$ gives two products $Q$ and $R$ each in $40 \,\%$ yield. 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 $....\, mg$.

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