Question

4 kg of 75% pure limestone is reacted as follows CaCO3+2HCl--CaCl2+CO2+H2O, volume of C02

Answer 1

Answer: Volume of carbon dioxide gas produced is 668.6 L.

Explanation:

Chemical reaction:

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

4 kg of 75% pure limestone means 3 Kg of limestone reacted.

$\text{Moles of limestone}=\frac{3000 g}{100 g/mol}=30 mol$

According to reaction one mole of limestone reacts to give one mole of carbon-dioxide.

Then 30 moles of limestone will give 30 moles of carbon-dioxide gas.

$\text{Amount of substance}=\text{number of moles of substance}\times \text{molecular mass of that substance}$

Volume of carbon dioxide gas produced:

Mass of carbon-dioxide produced$=30 \times 44 g/mol= 1320 g=1.324 Kg$         (1 Kg=1000 g)

Density of carbon-dioxide gas  = 1.98$Kg/m^3$

Volume of carbon-dioxide produced=$\frac{\text{mass of}CO_2}{\text{density of} CO_2}=\frac{1.324Kg}{1.98Kg/m^3}=0.6686 m^3=668.6 L (1m^3=1000L)$    

So, volume of carbon dioxide gas produced is 668.6 L.