Question

what volume of co2 at stp will evolve when 1g of caco3 reacts with excess of hcl​

Answer 1

Answer: 0.224 L

Explanation:

To calculate the moles,we use the following equation:  

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

$\text{Number of moles} of CaCO_3=\frac{1g}{100g/mol}=0.01moles$

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

$CaCO_3$ is considered as a limiting reagent because it limits the formation of product. As, the HCl is present in excess. Hence, it is considered as the excess reagent.

By Stoichiometry:  1 mole of $CaCO_3$ produces 1 mole of $CO_2$

So, 0.01 moles  $CaCO_3$ produces $=\frac{1}{1}\times 0.01=0.01moles$ of $CO_2$

According to Avogadro's law, 1 mole of every gas occupies 22.4 L at STP.

Thus 0.01 moles will occupy=$\frac{22.4}{1}\times 0.01=0.224L$

Answer 2

1. find mole of caco3
2. write balanced equation of caco3 with hcl
3. then find moles of co2 formed
4. find volume of co2 after finding mole