Question

If 20 g of CaCO3 is treated with 20 gram of HCl, how many grams of CO2 can be generated according to the given equation: CaCO3*(s)* + 2HCl *(aq)* ---------- CaCl2 *(aq)* + H2O*(l)* + CO2*(g)

Answer 1

Generated mass of $\bold{\mathrm{CO}_{2}}$ during in the given reaction is 8.8 g.

Explanation:

Balanced reaction:  $\mathrm{CaCO}_{3}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$

From the reaction,  

1 mole of calcium carbonate reacts with 2 moles of Hydrogen chloride and produce 1 mole of carbon dioxide.

Let’s calculate the moles of $\mathrm{CaCO}_{3}$¬:

Molar mass of $\mathrm{CaCO}_{3}$ = 100 g/mol

Given mass of $\mathrm{CaCO}_{3}$= 20 g

Moles of $\mathrm{CaCO}_{3}=\frac{\text { Mass of } \mathrm{CaCo}_{3}}{\text { Molar mass of } \mathrm{CaCo}_{3}}=\frac{20}{100}=0.20\ \mathrm{mol}$

0.20 mol of $\bold{\mathrm{CaCO}_{3}}$ reacts with 0.4 mol of HCl

Let’s calculate the moles of HCl:

Molar mass of HCl = 36.45 g/mol

Given mass = 20 ml

Moles of HCl= $\frac{\text { Mass of }\mathrm{HCl}}{\text {Molar mass of }\mathrm{HCl}}=\frac{20}{36.45}=0.548\ \mathrm{mol}$

Therefore, $\mathrm{CaCO}_{3}$ is limiting reagent.

0.20 mol of $\mathrm{CaCO}_{3}$ gives 0.20 mol of $\mathrm{CO}_{2}$

Molar mass of $\mathrm{CO}_{2}$ = 44 g/mol

Mass of $\mathrm{CO}_{2}\ produced = 0.2 \times 44 \frac{g}{m o l}=8.8 g$

Answer 2

Answer:

Here the moles of CaCO3 = Mass of CaCO3/

Molecular Mass

n= 20/100

n= 0.2

Moles of HCl = 20/36.5 = 0.54 mol

The reaction is  

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

For 2 mol of HCl, 1 mol CaCO3 is req.

This means for 0.54 mol HCl, 0.27 mol CaCO3 would be req.

But only 0.2 mol CaCO3 is present.

This means CaCO3 is Limiting Reagent.

For 1 mol CaCO3, 1 mol CO2 is formed.

Therefore moles of CO2 produced = 0.27

Mass of CO2 = 44 × 0.27

= 11.88 gm