100 g CaCO3 reacts with 1L,1M HCl. How much CO2 will be obtained on complete reaction

Asked by Anil | 16th May, 2017, 10:07: PM

Expert Answer:

Chemical reaction: CaCO3 + 2HCl → CaCl2  +  CO2 + H2O

Moles in 100 g CaCO3 = 100 g / 100 g mol-1 = 1 mol

Since molarity of a solution means the number of moles of solute in 1 litre of solution. The number of moles in 1 L, 1 M HCl = 1 mol

In the above reaction, for 1 mol of CaCO3 we require 2 mol of HCl, but we have only 1 mol of HCl is a limiting reactant.

As 2 moles of HCl gives 1 mol of CO2, 1 mol of HCl will produce = (1/2) x 1 = 0.5 mol of CO2.

Hence, on complete reaction, 0.5 mol or 22 g (0.5 mol x 44 g mol-1) of CO2 will be obtained.

Answered by Prachi Sawant | 17th May, 2017, 01:49: PM