🤓 Based on our data, we think this question is relevant for Professor Broering's class at CU.

Given reaction: MgCO_{3}(s) ⇌ MgO(s) + CO_{2}(g

Solve moles CO_{2}:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{n}}{\mathbf{=}}\frac{\mathbf{PV}}{\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{\left(\mathbf{0}\mathbf{.}\mathbf{0260}\mathbf{}\overline{)\mathbf{atm}}\right)\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{}\overline{)\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}\frac{\overline{)\mathbf{L}}\mathbf{\xb7}\overline{)\mathbf{atm}}}{\mathbf{mol}\mathbf{\xb7}\overline{)\mathbf{K}}}\mathbf{)}(650\overline{)K})}$

**nCO _{2} = 6.14 x 10^{-3 }mol CO_{2}**

Solve moles MgO:

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{}\mathbf{g}\mathbf{}\mathbf{MgO}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{MgO}}{\mathbf{40}\mathbf{.}\mathbf{3}\overline{)\mathbf{}\mathbf{g}\mathbf{}\mathbf{MgO}}}\mathbf{=}$** 0.0248 ^{ }mol MgO**

**ICE Chart: ***(reverse reaction during compression)*

At 650 K, the reaction MgCO_{3}(s) ⇌ MgO(s) + CO_{2}(g) has K_{p} = 0.026. A 12.6 L container at 650 K has 1.0 g of MgO(*s*) and CO_{2} at P = 0.0260 atm. The container is then compressed to a volume of 0.600 L. Find the mass of MgCO_{3} that is formed.

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Based on our data, we think this problem is relevant for Professor Broering's class at CU.