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

`\%\ mass\ of\ CaCO_3=93.37\ \%`

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.0493\ mol= (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=(Mass_(CaCO_3))/(Total\ mass)* 100`

`\%\ mass\ of\ CaCO_3=(4.93)/(5.28)* 100`

`\%\ mass\ of\ CaCO_3=93.37\ \%`