What volume of carbon (lV) oxide is produced at s.t.p. when 2.5g of CaCO3 reacts with excess acid according to the following equation? CaCO3 + 2HCI → CaCI2 +H2O + CO2 [CaCO3 = 100; molar volume of a gas at s.t.p = 22.4dm3]
CaCO3 + 2HCI → CaCI2 +H2O + CO2
Firstly, be sure that stoichiometrically, the equation is balanced. The ratio of CaCO3 to CO2 is 1:1
Secondly, what mass of CaCO3 will produce 22.4dm3 of CO2? (S.T.P)
To answer this question, calculate the molar mass of CaCO3 = 40 + 12 + (16X3) = 100g/mol
i.e 100g in 1 mol, which means the mass of CaCO3 that will produce 22.4 dm3 of CO2 at STP is 100g
If 100g of CaCO3 produces 22.4dm3 CO2
2.5g of CaCO3 will produce x dm3 CO2
Making x the subject of the relation, X = (2.5 X 22.4) / 100
x = 0.56dm3
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