Question

How many grams of carbon-di-oxide gas will be produced when 16.8 grams of 90% magnesium carbonate reacts with excess hydrochloric acid?​

Answer 1

Explanation:

The reaction is MgCO

3

+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.