417 g of PCl_{5(s)} is kept in an open conta | vedclass

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(B) The balanced chemical equation for the reaction is: $PCl_{5(s)} + H_2O_{(g)} \rightarrow POCl_{3(s)} + 2HCl_{(g)}$ Molar mass of $PCl_5 = 31 + 5 \

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$20$

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(B) The balanced chemical equation for the reaction is: $PCl_{5(s)} + H_2O_{(g)} ightarrow POCl_{3(s)} + 2HCl_{(g)}$ Molar mass of $PCl_5 = 31 + 5 	imes 35.5 = 208.5 \ g/mol$. Molar mass of $POCl_3 = 31 + 16 + 3 	imes 35.5 = 153.5 \ g/mol$. Let $x$ moles of $PCl_5$ react. Initial mass of $PCl_5 = 417 \ g$. Initial moles of $PCl_5 = 417 / 208.5 = 2 \ mol$. According to the reaction,$x$ moles of $PCl_5$ produce $x$ moles of $POCl_3$ and $2x$ moles of $HCl_{(g)}$. Since the container is open,$HCl_{(g)}$ escapes. The solid remaining consists of $(2 - x)$ moles of $PCl_5$ and $x$ moles of $POCl_3$. Mass of solid remaining $= (2 - x) 	imes 208.5 + x 	imes 153.5 = 395$. $417 - 208.5x + 153.5x = 395$. $417 - 55x = 395$. $55x = 22$. $x = 22 / 55 = 0.4 \ mol$. Percentage conversion $= (x / 2) 	imes 100 = (0.4 / 2) 	imes 100 = 20 \%$.

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