The volume of CO₂ at STP = 2.8 L
<h3>Further explanation</h3>
Given
21 gram of sodium hydrogen carbonate-NaHCO₃
Required
Volume of CO₂
Solution
The decomposition of sodium bicarbonate into sodium carbonate, carbon dioxide, and water :
<em>
2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)</em>
mol of NaHCO₃ :
= mass : MW NaHCO₃
= 21 g : 84 g/mol
= 0.25
From the equation, mol ratio of NaHCO₃(s) :CO₂(g) = 2 : 1, so mol CO₂ :
= 1/2 x mol NaHCO₃
= 1/2 x 0.25
= 0.125
At STP, 1 mol gas = 22.4 L, so for 0.125 mol :
= 0.125 x 22.4 L
= 2.8 L
