Q. A $0.60g$ sample consisting of only $MgC_{2}O_{4}$ and $CaC_{2}O_{4}$ is heated at $500^\circ C,$ converting the two salts to $MgCO_{3}$ and $CaCO_{3}.$
Weight of sample is $0.465g.$ What would the mixtures of oxides have weighed if the products were $CaO$ and $MgO$ and sample had been heated to $900^\circ C?$

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A

$0.12g$

B

$0.21g$

C

$0.25g$

D

$0.30g$

Solution:

Let $x$ g is wt. of $Ca_{2}O_{4}$ and $\left(\right.0.6-x\left.\right)g$ wt. of $MgC_{2}O_{4}$
$\mathrm{CaC}_2 \mathrm{O}_4 \stackrel{\Delta}{\rightarrow} \mathrm{CaCO}_3+\mathrm{CO}_2$
$\mathrm{MgC}_2 \mathrm{O}_4 \stackrel{\Delta}{\rightarrow} \mathrm{MgCO}_3+\mathrm{CO}_2$
wt. of $CaCO_{3}$ porduced $\text{=}\frac{x}{128}\times 100$
wt. of $MgCO_{3}$ produced $\text{=}\frac{\left(0 .6 - x\right)}{112}\times 84$
$\therefore \frac{x}{128}\times 100+\frac{\left(0 .6 - x\right)}{112}\times 84=0.465$
$\text{x}=0.48g$
Due to further heating
$CaCO_{3}\overset{\Delta }{ \rightarrow }CaO+CO_{2}$
$\frac{x}{128}$
$MgCO_{3}\overset{\Delta }{ \rightarrow }MgO+CO_{2}$
$\left(\frac{0 .6 - x}{112}\right)$
wt. of $CaO$ and $MgO$ produced
$=\frac{0 .48}{128}\times 56+\frac{0 .12}{112}\times 40=0.252g$

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