Question

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and 791 mmHg. Calculate the percent by mass of calcium carbonate in the sample.

Answer 1

Answer:

$\%\ mass\ of\ CaCO_3=93.37\ \%$

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of $CO_2$ produced =  0.0493 moles

According to the reaction:-

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

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

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,

$0.0493\ mol= \frac{Mass}{100.0869\ g/mol}$

$Mass_{CaCO_3}=4.93\ g$

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

$\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100$

$\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100$

$\%\ mass\ of\ CaCO_3=93.37\ \%$