Question

A $0.60\text{g}$ sample consisting of only $Ca{C}_2{O}_4$ and $Mg{C}_2{O}_4$ is heated at $500{}^{o}C$ converting the two salts to $CaC{O}_3$ and $MgC{O}_3$. The sample then weight $0.465\text{g}$. If the sample had been heated to $900{}^{o}C$, where the products are $CaO$ and $MgO$, what would the mixtures of oxide have weighed (in grams)?
Report your answer upto two decimal only.

Answer 1

Let $x\text{g}$ is the weight of $Ca{C}_2{O}_4$ and $(0.6-x)\text{g}$ is the weight of $Mg{C}_2{O}_4$
$Ca{C}_2{O}_4\underrightarrow{\Delta }CaC{O}_3+CO$

$Mg{C}_2{O}_4\underrightarrow{\Delta }MgC{O}_3+CO$
$\text{Weight of}CaC{O}_3\text{produced}=\frac{x}{128}\times 100$

$\text{Weight of}MgC{O}_3\text{produced}=\frac{(0.6-x)}{112}\times 84$
$\therefore \frac{x}{128}\times 100+\frac{(0.6-x)}{112}\times 84=0.465$
$x=0.48\text{g}$
Due to further heating,
$$\begin{array}{ccccc}CaC{O}_3& \underrightarrow{\Delta }& CaO& +& C{O}_2\\ & & \frac{x}{128}& & \end{array}$$
$$\begin{array}{ccccc}MgC{O}_3& \underrightarrow{\Delta }& MgO& +& C{O}_2\\ & & \left(\frac{0.6-x}{112}\right)& & \end{array}$$
Weight of $CaO$ and $MgO$ produced,
$=\frac{0.48}{128}\times 56+\frac{0.12}{112}\times 40=0.252\text{g}$