The stoichiometric reaction of 516 g of dime | vedclass

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(D) The reaction is: $4(CH_3)_2SiCl_2 + 4H_2O \rightarrow ((CH_3)_2SiO)_4 + 8HCl$.First,calculate the molar mass of dimethyldichlorosilane $(CH_3)_2Si

Vedclass · Accepted Answer

$222$

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(D) The reaction is: $4(CH_3)_2SiCl_2 + 4H_2O ightarrow ((CH_3)_2SiO)_4 + 8HCl$.First,calculate the molar mass of dimethyldichlorosilane $(CH_3)_2SiCl_2$: $2(12) + 6(1) + 28 + 2(35.5) = 24 + 6 + 28 + 71 = 129 \ g \ mol^{-1}$.Number of moles of $(CH_3)_2SiCl_2 = \frac{516 \ g}{129 \ g \ mol^{-1}} = 4 \ mol$.According to the stoichiometry,$4 \ mol$ of $(CH_3)_2SiCl_2$ produces $1 \ mol$ of the tetrameric cyclic product $X$ $(((CH_3)_2SiO)_4)$.Molar mass of $X = 8(12) + 24(1) + 4(28) + 4(16) = 96 + 24 + 112 + 64 = 296 \ g \ mol^{-1}$.Theoretical yield of $X = 1 \ mol 	imes 296 \ g \ mol^{-1} = 296 \ g$.Given the percentage yield is $75 \%$,the actual weight of $X$ obtained = $296 \ g 	imes \frac{75}{100} = 222 \ g$.

The stoichiometric reaction of $516 \ g$ of dimethyldichlorosilane with water results in a tetrameric cyclic product $X$ in $75 \%$ yield. The weight (in $g$) of $X$ obtained is. . . . . . . [Use,molar mass $(g \ mol^{-1})$: $H=1, C=12, O=16, Si=28, Cl=35.5$]

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