For the following reaction scheme,percentage | vedclass

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(C) $1$. $Mg_2C_3 + 4H_2O \rightarrow 2Mg(OH)_2 + CH_3C \equiv CH$ (Propyne,$P$). Molar mass of $P = 40 \ g \ mol^{-1}$. Moles of $P = 4.0 \ g / 40 \

Vedclass · Accepted Answer

$1.62, 3.2$

Vedclass · Answer

(C) $1$. $Mg_2C_3 + 4H_2O ightarrow 2Mg(OH)_2 + CH_3C \equiv CH$ (Propyne,$P$). Molar mass of $P = 40 \ g \ mol^{-1}$. Moles of $P = 4.0 \ g / 40 \ g \ mol^{-1} = 0.1 \ mol$.$2$. $P \xrightarrow{NaNH_2, MeI, 75\%} CH_3-C \equiv C-CH_3$ $(Q)$. Moles of $Q = 0.1 	imes 0.75 = 0.075 \ mol$.$3$. $3Q \xrightarrow{red \ hot \ iron \ tube, 40\%} 	ext{Hexamethylbenzene} (R)$. Moles of $R = (0.075 / 3) 	imes 0.40 = 0.01 \ mol$. Molar mass of $R (C_{12}H_{18}) = 162 \ g \ mol^{-1}$. $x = 0.01 	imes 162 = 1.62 \ g$.$4$. $P \xrightarrow{Hg^{2+}/H^+, 100\%} CH_3COCH_3 (S)$. Moles of $S = 0.1 \ mol$.$5$. $S \xrightarrow{Ba(OH)_2, \Delta, 80\%} (CH_3)_2C=CHCOCH_3 (T)$. Moles of $T = 0.1 	imes 0.80 = 0.08 \ mol$.$6$. $T \xrightarrow{NaOCl, 80\%} (CH_3)_2C=CHCOOH (U)$. Moles of $U = 0.08 	imes 0.80 = 0.064 \ mol$. Molar mass of $U (C_5H_8O_2) = 100 \ g \ mol^{-1}$. $y = 0.064 	imes 100 = 6.4 \ g$. (Note: Given options suggest $y=3.2$,implying a stoichiometry factor or yield interpretation difference in the source; based on standard calculation,$x=1.62$).

For the following reaction scheme,percentage yields are given along the arrows. $x \ g$ and $y \ g$ are the masses of $R$ and $U$,respectively. (Molar mass of $H$,$C$,and $O$ are $1$,$12$,and $16 \ g \ mol^{-1}$ respectively). $(1)$ The value of $x$ is $(2)$ The value of $y$ is.

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