Question

If 20 g of calcium carbonate is treated
with 20.0 g of HCl, how many grams of CO2 can be produced
according to the reaction.
CaCO3 + 2HCl (aq) CaCl(aq) + H20 (l) + CO2(g)
​

Answer 1

Molar mass of CaCO₃ = 100 g/mol

Molar mass of HCl = 36.5 g/mol

No. of moles of CaCO₃ = 20/100 = 0.2

No. of moles of HCl = 20/36.5 ≅ 0.55

Now, in this given question, one mole of CaCO₃ reacts with two moles of HCl to give one mole of CaCl₂, one mole of water and one mole CO₂

So, 0.2 moles of CaCO₃ reacts with 0.4 moles of HCl to give 0.2 mole of CaCl₂, 0.2 mole of H₂O and 0.2 mole of CO₂

Now, mass of 0.2 mole of CO₂ = 0.2 x MM of CO₂ = 0.2 x 44 = 8.8g

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