Q. 20 g of CaC$O_3$ is reacted with 20g of HCl. Find amount of C$O_2$ formed. Identify the limiting reagent. Calculate the amount of substance which remained unreacted.

Dear Student,

According to the given balanced equation, the stiochiometry suggests that 1 mol of CaCO₃ requires to evolve 1 mol of CO₂.
But in the reaction when 1 mol of ​CaCO₃ reacts with 2 mol of HCl it give 1 mol of CO₂. Or can be said as that 100 g of ​CaCO₃ reacts with 73 g of HCl to produce 44 g of CO₂.
So, 20 g of ​CaCO₃ will react with $\frac{73}{100}\times 20=14.6gofHCl$
So the limiting reagent is ​CaCO₃, and the number of moles of ​CaCO₃ consumed will be 20/100 = 0.2 moles
and according stiochiometry 0.2 mol of CO₂ will be produced = 0.2 x 44 = 8.8 g

So, the amount of ​CO₂ will be produced is 8.8 g


Regards.