Chem...Mole calculations

1) 0.936 g of a natural limestone was heated strongly to give quick lime (CaO) and CO2. The volume of CO2 collected at r.t.p. was 200 cm³. Find
A) the mass of quick lime formed.
B) The percentage of CaCO3 in the limestone (89.1%)

A)
CaCO3 →CaO + CO2

Mole ratio CaO : CO2 = 1 : 1
No. of moles of CO2 formed = 200/24000 = 0.008333 mol
No. of moles of CaO formed = 0.008333 mol
Molar mass of CaO = 40.1 + 16 = 56.1 g/mol
Mass of CaO formed = 0.008333 x 56.1 = 0.4675 g

B)
Mole ratio CaCO3 : CO2 = 1 : 1
No. of moles of CO2 formed = 0.008333 mol
No. of moles of CaCO3 formed = 0.008333 mol
Molar mass of CaCO3 = 40.1 + 12 + 16x3 = 100.1 g/mol
Mass of CaCO3 in the sample = 0.008333 x 100.1 = 0.8341 g
% of CaCO3 in the sample = (0.8341/0.936) x 100% = 89.1%


2) Under strong heating, baking power (NaHCO3) produces Na2CO3, CO2 and steam. Find the volume (at r.t.p.) of CO2 formed when 1.4 g NaHCO3 is heated.

2NaHCO3 → Na2CO3 + H2O + CO2
Mole ratio NaHCO3 : CO2 = 2 : 1

Molar mass of NaHCO3 = 23 + 1 + 12 + 16x3 = 84 g/mol
No. of moles of NaHCO3 used = 1.4/84 = 1/60 mol
No. of moles of CO2 formed = (1/60) x (1/2) = 1/120 mol
Volume of CO2 formed = (1/120) x 24000 = 200 cm³