Question

1.84 gram mixture of CaCO3 and MgCO3 isl heated gave 0.96g. What is the % mgCo3 composition of mixture also gave chemical equation​

Answer 1

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Molar mass of CaCO3 = 100g/mol

Molar mass of MgCO3=84.3 g/mol

Let number of moles of CaCO3 and MgCO3 in the mixture is X & Y respectively .

So

100X+84.3 Y= 1.84.......................(1)

On heating the mixture...

CO2 goes off from the product but mixture of CaO and MgO remains in the solid form.

Molar mass of CaO= 56g/mole

Molar mass of MgO= 40.3 g/mole

From the equation it is clear that number of moles of MgO & CaO in product mixture is same as that of respective carbonates in the reactant mixture.

so

56X+40.3Y=0.96........................(2)

Solving equation 1 and 2 we get.

Y=0.01

X=0.009

Thus mass of CaCO3 in mixture =100 X 0.0098 =0.98 g

Mass of MgCO3 in mixture = 84.3 X 0.01= 0.86 g

Mass % of CaCO3 in mixture=

Mass % of MgCO3 in mixture =

\$frac{0.98}{1.84} \times 100 = 53.26$\%1.840.98×100=53.26%

100-53.26= 46.74%

Answer 2

Molar mass of CaCO3 = 100g/mol

Molar mass of MgCO3=84.3 g/mol

Let number of moles of CaCO3 and MgCO3 in the mixture is X & Y respectively .

So

100X+84.3 Y= 1.84.......................(1)

On heating the mixture...

CO2 goes off from the product but mixture of CaO and MgO remains in the solid form.

Molar mass of CaO= 56g/mole

Molar mass of MgO= 40.3 g/mole

From the equation it is clear that number of moles of MgO & CaO in product mixture is same as that of respective carbonates in the reactant mixture.

so

56X+40.3Y=0.96........................(2)

Solving equation 1 and 2 we get.

Y=0.01

X=0.009

Thus mass of CaCO3 in mixture =100 X 0.0098 =0.98 g

Mass of MgCO3 in mixture = 84.3 X 0.01= 0.86 g

Mass % of CaCO3 in mixture=

Mass % of MgCO3 in mixture =

$\frac{0.98}{1.84} \times 100 = 53.26$

$\frac0.981.84×100=53.26$

$%1.840.98×100=53.26%$

$100-53.26= 46.74%$