Phosphine [PH_{3(g)}] decomposes to produce v | vedclass

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Question

(C) The balanced chemical equation for the decomposition of phosphine is: $4PH_{3(g)} \rightarrow P_{4(g)} + 6H_{2(g)}$.According to the stoichiometry

Vedclass · Accepted Answer

$+75$

Vedclass · Answer

(C) The balanced chemical equation for the decomposition of phosphine is: $4PH_{3(g)} ightarrow P_{4(g)} + 6H_{2(g)}$.According to the stoichiometry,$4$ volumes of $PH_3$ produce $1 + 6 = 7$ volumes of gaseous products.For $100 \ mL$ of $PH_3$:Volume of $P_4 = \frac{1}{4} 	imes 100 = 25 \ mL$.Volume of $H_2 = \frac{6}{4} 	imes 100 = 150 \ mL$.Total final volume = $25 + 150 = 175 \ mL$.Change in volume = $175 \ mL - 100 \ mL = +75 \ mL$.

Phosphine $[PH_{3(g)}]$ decomposes to produce vapours of phosphorus $(P_4)$ and $H_2$ gas. What will be the change in volume when $100 \ mL$ of phosphine is decomposed?

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