14.2 g mixture of magnesium carbonate and calcium carbonate produces 3.36 litres of carbon di-oxide in STP. Calculate the percentage of magnesium carbonate in the mixture.
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The reaction is MMgCO+CaCO 3
→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 0.2×22.4=4.48 L at STP.
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