Question

A mixture of Na2CO3 and MgCO3 of mass 7.63 g is reacted with an excess of hydrochloric acid. The CO2 gas generated occupies a volume of 1.67 L at 1.24 atm and 26 degree C. From these data, calculate the percent composition by mass of Na2CO3 in the mixture

Answer 1

Answer:

58.6 % by mass of Na₂CO₃

Explanation:

This is the reaction:

Na₂CO₃  +  MgCO₃ +  4HCl  →  MgCl₂  +  2NaCl  + 2CO₂  +  2H₂O

Let's find out the moles of CO₂ produced, by the Ideal Gases Law

1.24 atm . 1.67 L = n . 0.082 . 299K

(1.24 atm . 1.67 L / 0.082 . 299K) = n

0.0844 moles = n

Ratio is 2:1, so 2 moles of dioxide were produced by 1 mol of sodium carbonate. Let's make a rule of three:

2 moles of CO₂ were produced by 1 mol of Na₂CO₃

Then, 0.0844 moles of Co₂ would beeen produced by (0.0844 .1)/2  = 0.0422 moles of Na₂CO₃.

Let's convert this moles into mass (mol . molar mass)

0.0422 mol . 106 g/mol = 4.47 g

Finally we can know the mass percent of sodium carbonate in the mixture

(Mass of compound /Total mass) . 100 → (4.47 g / 7.63g) . 100 = 58.6 %