Recall the Ideal Gas Law equation:

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

**Step 1. **mol reactant

molar mass of (NH_{4})_{2}CO_{3} = **96.09 g/mol**

$\mathbf{11}\mathbf{.}\mathbf{9}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\left({\mathrm{NH}}_{4}\right)}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}}{\mathbf{96}\mathbf{.}\mathbf{09}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}}}\mathbf{=}$**0.124 mol (NH _{4})_{2}CO**

**Step 2. **volume of each product

$\mathbf{PV}\mathbf{=}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\frac{\overline{)\mathbf{P}}\mathbf{V}}{\overline{)\mathbf{P}}}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{P}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{nRT}}{\mathbf{P}}}$

P = **1.03 atm**

T = 23.0˚C + 273.15 = **269.15 K**

R =** 0.08206 (L·atm)/(mol·K)**

Ammonium carbonate decomposes upon heating according to the following balanced equation: (NH_{4})_{2}CO_{3}(s) → 2 NH_{3}(g) + CO_{2}(g) + H_{2}O(g). Calculate the total volume of gas produced at 23.0˚C and 1.03 atm by the complete decomposition of 11.9 g of ammonium carbonate.

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