Calculate moles O_{2} produced using the ideal gas equation:

$\overline{)\mathbf{PV}\mathbf{=}\mathbf{nRT}}$

$\mathbf{P}\mathbf{=}\mathbf{746}\mathbf{}\overline{)\mathbf{mmHg}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{atm}}{\mathbf{760}\mathbf{}\overline{)\mathbf{mmHg}}}$

P = 0.9815 atm

V = 4.55 L

T = 308 K

$\frac{\mathbf{PV}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{RT}}}{\overline{)\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{(0.9815\overline{)\mathrm{atm}})(4.55\overline{)L})}{(0.08206{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}})(308\overline{)K})}$

In the reaction shown here, 4.55 L of O_{2} was formed at P = 746 mmHg and T = 308 K .

2Ag_{2}O(s) → 4Ag(s)+O_{2}(g)

How many grams of Ag_{2}O decomposed?

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