Question

A 6.875 gram sample of a mixture of Na₂CO₃(s) and NaI(s) is dissolved in 145 mL of distilled water. The Na⁺ molarity of the resulting solution is 0.577 M.

In another experiment 4.96 grams of the mixture is added to the 88.0 grams of 1.0 M HCl. The reaction is allowed to go to completion. The final mass after reaction is 92.03 grams.

Calculate the mass % of Na₂CO₃ and NaI in the mixture.

Answer 1

Solution :-

When HCl is added then it reacts with Na2CO3 by the folllwing reaction equation

Na2CO3 + 2HCl ------ > CO2 + H2O + 2NaCl

So after the reaction the CO2 escapes from the reaction mixture

Therefore lets calculate the mass of the CO2 escaped

Mass of CO2 = (mass of reactant + HCl solution ) – final mass

                       = (4.96 g + 88.0 g) – 92.03 g

                       = 0.93 g CO2

Now lets calculaete the moles of the CO2

Moles of CO2 = 0.93 g / 44.01 g per mol = 0.02113 mol CO2

Now using the mole ratio of the CO2 and Na2CO3 we can find moles of Na2CO3

0.02113 mol CO2 * 1 mol Na2CO3 / 1 mol CO2 = 0.02113 mol Na2CO3

Now lets convert moles of Na2CO3 to its mass

Mass = moles * molar mass

          = 0.02113 mol * 105.98 g per mol

          = 2.239 g Na2CO3

Now lets calculate the percent of the each

% of Na2CO3 = (mass of Na2CO3 / total mass )*100%

                          =( 2.239 g / 4.96 g)*100%

                          = 45.14 % Na2CO3

Now lets find the % of NaI

% of NaI = 100 % - % of Na2CO3

                 = 100 % - 45.14 %

                 = 54.86 % NaI