Question

3. A gas mixture has the following percentages by mass: N2: 70% O2: 14% CO: 4% CO2: 12% What are the mole percentages of the gases in the mixture?

Answer 1

Solution :-

Lets assume we have 100 g of the gas sample then mass of each gas is N2 = 70 g , O2 = 14 g, CO = 4 g and CO2 = 12 g

now lets calculate moles of each gas

formula

moles = mass / molar mass

moles of N2 = 70 g / 28.014 g per mol =2.499 mol N2

moles of O2 = 14 g / 32 g per mol = 0.4375 mol O2

moles of CO = 4 g / 28.0 g per mol =0.14286 mol CO

moles of CO2 = 12 g / 44.01 g per mol = 0.27267 mol CO2

now lets calculate the total moles of gases

total moles = 2.499 mol N2 + 0.4375 mol O2+0.14286 mol CO + 0.27267 mol CO2

                 =3.35203 mole gases

now lets calculate the mole percent of the each gas

mole % of gas = (moles of the gas / total moles of gases)*100%

mole % of N2 = (2.499 mol / 3.35203 mol)*100 %

                      = 74.55 %

mole % of O2 =(0.4375 mol / 3.35203 mol )*100%

                    = 13.05 %

mole % of CO = (0.14286 mol / 3.35203 mol )*100%

                       = 4.26 %

mole % of CO2 = (0.27267 mol / 3.35203 mol)*100%

                        = 8.13 %