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On heating 0.69 g sample of $\mathrm{CaC}_2 \mathrm{O}_4$ & $\mathrm{MgC}_2 \mathrm{O}_4$ at a temperature of $300^{\circ} \mathrm{C}, \mathrm{CaCO}_3\; \&\; \mathrm{MgCO}_3$ were formed respectively. Resulting sample was 0.53 g. On further heating at $700^{\circ} \mathrm{C}, \mathrm{CaO}\; \& \; \mathrm{MgO}$ were formed.

Calculate the wight of the oxides so formed in the mixture.
Option: 1
0.362

Option: 2
1.088

Option: 3
0.143

Option: 4
0.275

Answers (1)

best_answer

Let,

$\mathrm{m_{\mathrm{MgC}_2 \mathrm{O}_4}=0.69-x}$

$\mathrm{m_{\mathrm{CaC}_2 \mathrm{O}_4}=x}$

$\begin{aligned} & \mathrm{CaC}_2 \mathrm{O}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{CaCO}_3+\mathrm{CO} \\ \\& \mathrm{MgC}_2 \mathrm{O}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{MgCO}_3+\mathrm{CO} \end{aligned}$

Weight of $\mathrm{CaCO_{3}}$ formed $=\mathrm{\frac{x}{128}\times 100}$

Weight of $\mathrm{MgCO_{3}}$ formed $=\mathrm{\frac{0.69-x }{112}\times 84}$

$\mathrm{\frac{x}{128} \times 100+\frac{0.69-x}{112} \times 84=0.53}$

$\mathrm{x=0.40 \mathrm{~g}}$

On further heating

$\mathrm{\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2}$

Moles of $\mathrm{ \mathrm{CaO}=\frac{x}{128}}$

$\mathrm{MgCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_2$

Moles of $\mathrm{M g O=\frac{0.69-x}{112}}$

Weight of $\mathrm{\mathrm{CaO} \; \&\; \mathrm{MgO}=\frac{0.40}{128} \times 56+\frac{0.28}{112} \times 40}$

$=0.175+0.1=0.275$

Posted by

Ritika Kankaria

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