Recall the Ideal Gas Law equation:

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

**Step 1. **mol reactant

molar mass of NH_{4}NO_{3} = **80.043 g/mol**

$\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{}\overline{)\mathbf{kg}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\overline{)\mathbf{g}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{kg}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}}{\mathbf{96}\mathbf{.}\mathbf{09}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{NO}}_{\mathbf{3}}}}\mathbf{=}$**16.2 mol**

**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 = 751 mmHg** = 0.988 atm**

T = 126˚C + 273.15 = **399****.15 K**

R =** 0.08206 (L·atm)/(mol·K)**

Ammonium nitrate decomposes explosively upon heating according to the following balanced equation: 2 NH_{4}NO_{3}(s) → 2 N_{2}(g) + O_{2}(g) + 4 H_{2}O(g). Calculate the total volume of gas (at 126˚C and 751 mmHg) produced by the complete decomposition of 1.56 kg of ammonium nitrate.

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