Chemistry - Mole Concept Question with Solution | TestHub

$\begin{array}{ccccccc}{\text{MgCO}}_3& +& {\text{CaCO}}_3& ⟶& \text{MgO}+\text{CaO}& +& 2{\text{CO}}_2\left(\text{g}\right)\\ \left(1.84-\text{x}\right)\text{g}& & \text{x}\text{g}& & & & 0.88\text{g}\\ \text{Moles}\{\frac{1·84-\text{x}}{84}& & \frac{\text{x}}{100}& & & & \frac{0·88}{44}\end{array}$
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Apply$\text{POAC}$to$\text{C}$

$\frac{1.84-\text{x}}{84}\times 1+\frac{\text{x}}{100}\times 1=\frac{0.88}{44}\times 1=0.02$

$184-100\text{ x}+84\text{ x}=168$

$\therefore$$16=16\text{ x}$

$\therefore$$\text{x}=1\text{ g}$