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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2
Relative molecular mass of NaHCO3 = 23 +1 + 12 + 16 * 3 = 84smol-1
2 * 84g of NaHCO3 will be liberated 22.4dm3 or CO2 at s.t.p
16.8g of NaHCO3 will libertae (16.8 * 22.4)/(2 * 84) = 2.24 dm3