Question

A sample contained a mixture of BaCO3 and MgCO3. The amount of BaCO3 was determined by reacting 10.53 g of the sample with an excess of HCl to release CO2 from each compound; BaCO3(s) + 2 HCl(aq) --> BaCl2 + CO2(g)

MgCO3(s) + 2 HCl(aq) --> MgCl2 + CO2(g)

If the sample reacted completely and produced 2230 mL of carbon dioxide at 30.86 °C and 738.2 mm Hg, what was the mass percentage of BaCO3 in the mixture?

Answer 1

Suppose mass of BaCO3 in mixture = x g then mass of MgCO3 = (10.53 – x) g

Molar mass of BaCO3 = 197.34 g/mole, Molar mass of MgCO3 = 84.314 g/mole

Number of moles of BaCO3 = x g / 197.34 g/mole =(x/197.34) mole

Number of moles of MgCO3 = (10.53-x) g / 84.314 g/mole = (10.53-x/84.314) mole

From reaction, 1.0 mole of BaCO3 produces 1.0 mole of CO2, hence (x/197.34) mole of BaCO3 will produce (x/197.34) mole of CO2

From reaction, 1.0 mole of MgCO3 produces 1.0 mole of CO2, hence (10.53-x/84.314) mole of MgCO3 will produce (10.53-x/84.314) mole of CO2

Total number of moles of CO2 produced = {(x/197.34) + (10.53-x/84.314)} mole

Given,

Volume of CO2 = 2230 ml = 2.23 L, Temperature = 30.86 ^oC = (30.86 +273.15) = 304.01 K

Pressure = 738.2 mmHg = (738.2 / 760) atm = 0.971 atm

We know the relationship from gas law, number of moles(n) = PV / RT

Number of moles of CO2 produced = (0.971 atm * 2.23 L) / (0.082 atm L mol^-1K^-1 *304.1 K)

Number of moles of CO2 produced = 2.16533 mole / 24.9362 = 0.087 mole

Now we can write

{(x/197.34) + (10.53-x/84.314)} mole = 0.087 mole

84.314 x + 197.34 (10.53 – x) = 197.34 *84.314 *0.087

84.314 x + 2077.99 – 197.34 x = 1447.55

-113.026 x = -630.44

x = 630.44 / 113.026 = 5.578

Mass of BaCO3 in mixture = x g = 5.578 g

% BaCO3 in the mixture = (5.578 g / 10.53 g)*100 = 52.97%