When a solid mixture of mgco3 and caco3 is heated strongly, carbon dioxide gas is given off and a solid mixture of mgo and cao is obtained. if a 24.00 g sample of a mixture of mgco3 and caco3 produces 12.00 g co2, then what is the percentage by mass of mgco3 in the original mixture? express your answer u?

MgCO3---> MgO + CO2, x mol MgCO3

M (MgCO3) = 24.3+12.0+48.0=84.3 g/mol

CaCO3--->CaO+CO2, y mol CaCO3

M (CaCO3) = 40.0+12.0+48.0=100g/mol

M (CO2) = 44.0

x*84.3 + y*100=24.00

x*44+y*44=12.00 x+y=12/44=0.2727, x=0.2727-y

(0.2727-y) * 84.3 + y*100=24.00

22.99-84.3y+100y=24.00

22.99+15.7y = 24.00, 15.7y=1.01, y=0.06430 mol CaCO3

x=0.2727-0.06430=0.2084 mol MgCO3

0.06430 mol CaCO3*100g/mol=6.43 g CaCO3

0.2084 mol MgCO3*84.3g/mol=17.57 g MgCO3

what does it mean express your answer u?