Question

What volume of co2 at ntp be obtained on treating 1gram of caco3 with dilute hcl marble

Answer 1

M.mass of CaCO3 is 100g
CaCO3 + 2HCl gives CO2+ CaCl2 + H2O
100g CaCO3 gives 22.4L CO2
So 1g gives 22.4÷100 = 22.4 × 10^-2

Answer 2

Answer: 0.2404 L of $CO_2$ will be obtained at NTP.

Explanation: When calcium carbonate reacts with dilute hydrochloric acid, reaction follows:

$CaCO_3(s)+2HCl(aq.)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)$

Number of moles can be calculated by:

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

$\text{Molar mass of }CaCO_3=100g/mol$

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

By Stoichiometry,

1 mole of calcium carbonate produces 1 mole of carbon dioxide.

$\text{0.01 moles of calcium carbonate will produce}=\frac{1}{1}\times 0.01=0.01$

At NTP, the conditions are:

P = 1 atm

T = 20 °C = (273 + 20) = 293 K

$R=0.082057\text{ L atm }mol^{-1}K^{-1}$

n = 0.01 moles (Calculated above)

Using Ideal gas equation, we get

PV = nRT

Putting values in above equation, we get

$V=\frac{(0.01mol)(0.082057\text{ L atm }mol^{-1}K^{-1})(293K)}{(1atm)}$

Volume = 0.2404 L