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Ideal Gas Equation

What total vo...

Question

What total volume, in litre at $627^{o}$ C and $1$ atm, could be formed by the decomposition of $16$ gm of $N H_{4} N O_{3}$ ?
Reaction: $2 N H_{4} N O_{3} \to 2 N_{2} + O_{2} + 4 H_{2} O (g)$ .

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Solution

$N H_{4} N O_{3} \to N_{2} O + 2 H_{2} O$
$\frac{(16 g - N H_{4} N O_{3})}{(80.0434 g - N H_{4} N O_{3} / m o l)} \times \frac{(3 m o l - g a s e s)}{(1 m o l - N H_{4} N O_{3})}$ = 0.59967 mol gases
$V = \frac{n R T}{P}$ = $(0.59967 m o l) \times (0.08205746 L a t m / K m o l) \times (627 + 273) K / (1 a t m)$ =
$44 L$ water vapor and nitrous oxide

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