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Question

$$18.4$$ g of a mixture of calcium carbonate and magnesium carbonate, on heating, gives $$4.0$$ g of magnesium oxide. The volume of $$CO_{2}$$ produced at STP in this process is:


A
1.12 L
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B
4.48 L
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C
2.24 L
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D
3.36 L
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Solution

The correct option is B $$4.48$$ L
The reaction is $$MgCO_{3}+CaCO_{3}\rightarrow MgO+CaO+2CO_{2}$$.
$$1$$ mole of magnesium carbonate (molecular weight $$84$$ g/mol) gives $$1$$ mole of magnesium oxide (molecular weight is $$40$$ g/mol).
Thus, $$4.0$$ g of magnesium oxide is obtained from $$8.4$$ g ($$0.1$$ mol) of magnesium carbonate.
Thus, in $$18.4$$ g of mixture, $$18.4-8.4=10$$ g ($$0.1$$ mol) of calcium carbonate (molecular weight $$=100$$ g/mole) is present.
A mixture of $$1$$ mol of magnesium carbonate and $$1$$ mole of calcium carbonate gives $$2$$ moles of carbon dioxide.
Hence, a mixture of $$0.1$$ mole of magnesium carbonate and $$0.1$$ mole of calcium carbonate will give $$0.2$$ moles of carbon dioxide.
At STP, $$1$$ mole of carbon dioxide occupies $$22.4$$ L.
Hence, $$0.2$$ mole of carbon dioxide will occupy $$0.2 \times 22.4 =4.48$$ L at STP.

Chemistry

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