For the given reaction; CaCO3 + 2HCl CaCl2 + H2O + CO2 If 90 g CaCO3 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 1.13 g mL-1, then which of the following option is correct? Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol-1 respectively.
- 64.97 g of HCl remains unreacted
- 60.32 g of HCl remains unreacted
- 32.85 g of CaCO3 remains unreacted
- 97.30 g of HCl reacted
Answer
(1) 64.97 g of HCl remains unreacted
Solution
Mass of HCl solution = 300 1.13 = 339 g
Mass of HCl = 339 38.55100 = 130.68 g
nHCl = 130.6836.5 = 3.58 mol
nCaCO3 = 90100 = 0.9 mol (molar mass = 40 + 12 + 48 = 100)
1 mol CaCO3 needs 2 mol HCl, 0.9 mol needs 1.8 mol HCl (< 3.58 mol)
CaCO3 is the limiting reagent and is used up completely.
HCl reacted = 1.8 36.5 = 65.70 g
HCl unreacted = 130.68 - 65.70 = 64.98 g 64.97 g
Answer: (1)