Question

When solid calcium carbonate is reacted with aqueous hydrochloric acid, the products of the reaction include aqueous calcium chloride, liquid water, and gaseous carbon dioxide. Calculate the volume of CO₂ gas collected over water at 25.0 °C when 25.1 g of calcium carbonate is added to excess hydrochloric acid if the total pressure is 911 mm Hg. The vapor pressure of water at 25.0 °C is 23.8 mm Hg.

Answer 1

Answer:

V = 5.25 L

Explanation:

Vapor pressure of water = 23.8 mmHg

Total vapor pressure = 911 mmHg

Vapor pressure of carbon dioxide gas = Total vapor pressure - Vapor pressure of water = (911 - 23.8) mmHg = 887.2 mmHg

Mass of calcium carbonate = 25.1 g

Molar mass of $CaCO_3$ = 100.0869 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{25.1\ g}{100.0869\ g/mol}$

Moles of $CaCO_3$ = 0.2508 moles

From the reaction shown below as;

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

1 mole of calcium carbonate on reaction forms 1 mole of carbon dioxide

0.2508 mole of calcium carbonate on reaction forms 0.2508 mole of carbon dioxide

Mole of carbon dioxide = 0.2508 moles

We use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 887.2 mmHg  

V = ?

T = Temperature of the gas = $25^oC=[25+273]K=298K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of carbon dioxide gas = 0.2508 moles

Putting values in above equation, we get:

$887.2mmHg\times V=0.2508 moles\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K$

V = 5.25 L