Question

The volume of $C{O}_2$ obtained at STP by the decompostion of 10 g of calcium carbonate is:
(molar mass of calcium carbonate = 100 $gmo{l}^{-1}$

• A. 2.24 L
• B. 44.8 L
• C. 67.2 L
• D. 22.4 L

Answer 1

The correct option is A 2.24 L

The balanced chemical equation is:
$CaC{O}_3\to CaO+C{O}_2$

Given mass of calcium carbonate = 10 g
Moles of calcium carbonate = $\frac{10}{100}=0.1$
Moles of $C{O}_2$ that can be obtained according to stoichiometry = 0.1
As, 1 mole of $C{O}_2$ corresponds to 22.4 L at STP.
$\therefore$ 0.1 mole of $C{O}_2$ corresponds to 2.24 L at STP