Question

A mixture of magnesium carbonate and barium carbonate with a total mass of 1.000 g was treated with excess of hydrochloric acid to produce 0.407 g of carbon dioxide gas. Write balanced reaction equations occurring to this system. Calculate the mass percent of magnesium carbonate in the starting mixture (with three significant figures).

Answer 1

MgCO3 + BaCO3 = 1 g

HCl addition

m = 0.407 g of CO2

find reaction:

MgCO3 +2HCl = MgCl2 + H2O + CO2

BaCO3 +2HCl = BaCl2 + H2O + CO2

this is already balanced

find mass %

mol of CO2 = mass/MW = 0.407/44 = 0.00925 mol of CO2

then

1:1 ratio

mass of BaCO3 + mass of MgCO3 = 1

mol of BaCO3 * MW + mass of MgCO3 = 1

mol of BaCO3*197.3359 + mol of MgCO3 * 84.3139 = 1

and we know

mol of BaCO3 + mol of MgCO3 = moles of CO2

Equations:

mol of BaCO3 + mol of MgCO3 =0.00925

mol of BaCO3*197.3359 + mol of MgCO3 * 84.3139 = 1

let "B" be BaCO3 and "M" be MgCO3

b + m = 0.00925

197.3359 b + 84.3139 m = 1

solve

m = 0.00925 -b

197.3359*b + 84.3139 *(0.00925 -b) = 1

197.3359*b + 84.3139 *(0.00925 -84.3139 b = 1

( 197.3359-84.3139 )*b = 1 -  84.3139 *(0.00925

b =  (1 -  84.3139 *(0.00925) ) /(197.3359-84.3139) = 0.00194 mol of BaCO3

mol of m = 0.00925 -0.00194 = 0.00731

mass of MgCO3 = 0.00731*84.3139 = 0.61633 g

% MgCO3 = mas MgCO3/Total mass * 100 %= 0.61633 / 1 *100 = 61.6 %