Question

A $0.60g$ sample consisting of only $Ca{C}_2{O}_4$ and $Mg{C}_2{O}_4$ is heated at ${500}^{\circ }C$, converting the two salts of $CaC{O}_3$ and $MgC{O}_3$. The sample then, weighs $0.465g$. If the sample had been heated to ${900}^{\circ }C$, where the products are $CaO$ and $MgO$, what would the mixtures of oxides have weighed?

• A. $0.12g$
• B. $0.21g$
• C. $0.252g$
• D. $0.3g$

Answer 1

The correct option is C $0.252g$

Let $xg$ is the weight of $Ca{C}_2{O}_4$ and $(0.6-x)$ g be the weight of $Mg{C}_2{O}_4$.

The reactions are as follows:
$Ca{C}_2{O}_4\stackrel{△}{\to }CaC{O}_3+C{O}_2$
$Mg{C}_2{O}_4\stackrel{△}{\to }MgC{O}_3+C{O}_2$

Weight of $CaC{O}_3$ produced$=\frac{x}{128}\times 100$.

Weight of $MgC{O}_3$ produced$=\frac{(0.6-x)}{112}\times 84$.

$\therefore \frac{x}{128}\times 100+\frac{(0.6-x)}{112}\times 84=0.465$.

Therefore, $x=0.48g$

Due to further heating,
$CaC{O}_3\stackrel{△}{\to }CaO+C{O}_2$

- $\frac{x}{128}$

$MgC{O}_3\stackrel{△}{\to }MgO+C{O}_2$

- $\frac{0.6-x}{112}$

Weight of $CaO$ and $MgO$ produced $=\frac{0.48}{128}\times 56+\frac{0.12}{112}\times 40=0.252g$.

Hence, option C is correct.