Question

84 grams mixture of calcium carbonate and magnesium carbonate upon heating air 0.96 gram residue calculate the percentage composition of the mixture

Answer 1

your question is -> 1.84 gram mixture of CaCO3 and MgCO3 is strongly heated till no further loss in mass is observed. The mass of residue is 0.96g. What is the % age composition of mixture?

solution : Let mass of MgCO3 in mixture is x and then mass of CaCO3 is (1.84 - x)

now, mole of MgCO3 = x/(24 + 12 + 48) = x/84

mole of CaCO3 = (1.84 - x)/(40 + 12 + 60) = (1.84 - x)/100

1.84 g of mixture of CaCO3 and MgCO3 is strongly heated till no further loss in mass is observed.

i.e., chemical equations are ....

$CaCO_3\rightarrow CaO + CO_2(\uparrow)$

$MgCO_3\rightarrow MgO+CO_2(\uparrow)$

1 mole of MgCO3 residues one mole of MgO .

similarly, one mole of CaCO3 residues one mole of CaCO3.

so, x/84 mole of MgCO3 residues x/84 mole of MgO

and (1.84 - x)/100 mole of CaCO3 residues (1.84 - x)/100 of CO2.

now, total mass of MgO + mass of CaO = 0.96

or, x/84 × (24 + 16) + (1.84 - x)/100 × (40 + 16) = 0.96

or, 40x/84 + 56(1.84 - x)/100 = 0.96

hence x ≈ 0.84 g and then (100 - x) = 1.84 - 1 = 1g

so, percentage of MgCO3 = 0.84/1.84 × 100 = 45.7 %

percentage of CaCO3 = 1/1.84 × 100

= 54.3 %