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Question
Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide were collected at 27oC and normal pressure.
\[\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}\]
Calculate the mass of salt required.
Solution 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
Solution 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
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