Question

Pure calcium carbonate and dilute hydrochloric acid reacted and 2 litres of carbon dioxide was collected at 27°C and normal pressure.CaCO3 + 2HCl —> CaCl2 +H2O + CO2Calculate :(a) the mass of salt required(b) the mass of the acid required​

Answer 1

Answer :

(a) The mass of salt required is, 8.12 grams

(b) The mass of the acid required is, 5.93 grams

Explanation : Given,

Volume of carbon dioxide gas = 2 L

Temperature of carbon dioxide gas = $27^oC=273+27=300K$

Pressure of carbon dioxide gas = 1 atm

First we have to calculate the moles of carbon dioxide gas by using 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

$(1atm)\times (2L)=n\times (0.0821L.atm/mole.K)\times (300K)$

$n=0.0812mole$

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

The balanced chemical reaction is,

$CaCO_3(s)+2HCl(l)\rightarrow CaCl_2(s)+H_2o(l)+CO_2(g)$

From the balanced reaction we conclude that,

As, 1 mole of $CO_2$ obtained from 1 mole of $CaCO_3$

So, 0.0812 mole of $CO_2$ obtained from 0.0812 mole of $CaCO_3$

$\text{Mass of }CaCO_3=\text{Moles of }CaCO_3\times \text{Molar mass of }CaCO_3=0.0812mole\times 100g/mole=8.12g$

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

(b) Now we have to calculate the mass of acid required.

The balanced chemical reaction is,

$CaCO_3(s)+2HCl(l)\rightarrow CaCl_2(s)+H_2o(l)+CO_2(g)$

From the balanced reaction we conclude that,

As, 1 mole of $CO_2$ obtained from 2 mole of $HCl$

So, 0.0812 mole of $CO_2$ obtained from 0.1624 mole of $HCl$

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl=0.1624mole\times 36.5g/mole=5.93g$

Therefore, the mass of the acid required is, 5.93 grams

Answer 2

Explanation:

Same as stated above

Thanks