Question

60 grams of limestone on heating produce 22 grams of CO2. The percentage of calcium carbonate in limestone is
please answer this...........

Answer 1

CaCO3 = CaO + CO2

100 g 44 g

60 g 44 x 60/100 = 26.4 g

60 g of 100% pure limestone would give 26.4 g of CO2

But we got only 22 g

So Purity = 22 x 100/26.4

= 83.33%

Answer 2

Answer: The percentage of calcium carbonate in limestone is 83.33 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of carbon dioxide = 22 g

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon dioxide}=\frac{22g}{44g/mol}=0.5mol$

The chemical equation for the heating of limestone follows:

$CaCO_3\rightarrow CaO+CO_2$

By Stoichiometry of the reaction:

1 mole of carbon dioxide is produced by 1 mole of calcium carbonate

Now, calculating the mass of calcium carbonate from equation 1, we get:

Molar mass of calcium carbonate = 100 g/mol

Moles of calcium carbonate = 0.5 moles

Putting values in equation 1, we get:

$0.5mol=\frac{\text{Mass of calcium carbonate}}{100g/mol}\\\\\text{Mass of calcium carbonate}=(0.5mol\times 100g/mol)=50g$

To calculate the percentage of calcium carbonate in limestone, we use the equation:

$\%\text{ composition of calcium carbonate}=\frac{\text{Mass of calcium carbonate}}{\text{Mass of limestone}}\times 100$

Mass of calcium carbonate = 50 g

Mass of limestone = 60 g

Putting values in above equation, we get:

$\%\text{ composition of calcium carbonate}=\frac{50}{60}\times 100\\\\\%\text{ composition of calcium carbonate}=83.33\%$

Hence, the percentage of calcium carbonate in limestone is 83.33 %