Question

Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide were collected at 27^oC and normal pressure.

\[\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}\]

Calculate the mass of salt required.

Answer 1

 Given,

The volume of carbon dioxide gas = 2 L

The temperature of carbon dioxide gas = 27°C = 273 + 27 = 300K

The pressure of carbon dioxide gas = 1 atm

First, we have to calculate the moles of carbon dioxide gas by using the ideal gas equation.

PV = nRT

where,

P = pressure of carbon dioxide gas

V = volume of carbon dioxide gas

T = temperature of carbon dioxide gas

n = number of moles of carbon dioxide gas

R = gas constant = 0.0821 L.atm/mole.K

(1 atm) × (2L) = n × (0.0821 L.atm/mole.K) × (300K)

n = 0.0812 mole

(a) Now we have to calculate the mass of salt required.

The balanced chemical reaction is,

\[\ce{CaCO3(s) + 2HCl(l) -> CaCl2(s) + H2O(l) + CO2(g)}\]

From the balanced reaction, we conclude that,

As, 1 mole of CO₂ obtained from 1 mole of CaCO₃ 

So, 0.0812 moles of CO₂ obtained from 0.0812 moles of CaCO₃

Mass of CaCO₃ = Moles of CaCO₃ × Molar mass of CaCO₃ = 0.0812 mole × 100g/mole = 8.12 g

Therefore, the mass of the salt required is, 8.12 grams

Answer 2

Given: 

V₁ = 2 litres                            V₂ = ?

T₁ = (273 + 27) = 300°K         T₂ = 273°K

As we know, `"V"_1/"T"_1 = "V"_2/"T"_2`

∴ V2 = `("V"_1"T"_2)/"T"_1`

= `[[2 × 273]/300]`L

Now at S.T.P. 22.4 litres of CO₂ are produced using CaCO₃ = 100 g

∴ `[[2 × 273]/300]` litres are produced by

= `100/22.4 xx (2 xx 273)/300`

= 8.125 g