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A 0.6 gm sample consisting of only ${CaC}_{2} O_{4}$ and ${MgC}_{2} O_{4}$ is heated at 500°C, gets converted into ${CaCO}_{3}$ and ${MgCO}_{3}$ . The sample then weighed 0.465 gm. If the sample had been heated to 900°C where the products are CaO and MgO. then what would the mixture of oxides weigh?

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0.12 gm

0.21 gm

0.252 gm

0.3 gm

answer is C.

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Detailed Solution

$\underset{a}{{CaC}_{2} O_{4}} \to \underset{a}{{CaCO}_{3}} + \underset{a}{CO}$

$\underset{b}{{MgC}_{2} O_{4}} \to \underset{b}{{MgCO}_{3}} + \underset{b}{CO}$

$\begin{array}{l} (a + b) \times 28 = 0.6 - 0.465 \\ \Rightarrow (a + b) = 4.82 \times 10^{- 3} \end{array}$

Loss in weight, due to ${CO}_{2}$

$\Rightarrow (a + b) \times 44 = 0.212 g$

New weight = 0.465 - 0.212 = 0.253 g

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