3.68 of a mixture of CaCO 3 and MgCO 3 is heated to liberate 0.04 mole of CO 2. The mole % of CaCO 3 and MgCO 3 in the mixture is respectively:

Question

3.68 of a mixture of CaCO 3 and MgCO 3 is heated to liberate 0.04 mole of CO 2. The mole % of CaCO 3 and MgCO 3 in the mixture is respectively: (A) 50 %, 50 % (B) 60 %, 40 % (C) 40 %, 60 % (D) 30 %, 70 %

Tardigrade Admin · Accepted Answer

Let the mass of CaCO 3= x g Then, mass of MgCO 2=(3.68- x ) g moles of CaCO 3=( x /100) moles of MgCO 3=(3.68- x /84) Applying POAC for C-atoms (x/100)+(3.68-x/84)=0,04 ∴ n CaCO 3=(2/100)=0.02 and n MgCO3=(1.68/84)=0.02 mole % of CaCO 3=(0.02/0.04) × 100=50 % mole % of MgCO 3=(0.02/0.04) × 100=50 %

Q. 3.68 of a mixture of $C a C O_{3}$ and $M g C O_{3}$ is heated to liberate 0.04 mole of $C O_{2}$ . The mole $%$ of $C a C O_{3}$ and $M g C O_{3}$ in the mixture is respectively:

$50%, 50%$

$60%, 40%$

$40%, 60%$

$30%, 70%$

Q. 3.68 of a mixture of CaCO3​ and MgCO3​ is heated to liberate 0.04 mole of CO2​. The mole % of CaCO3​ and MgCO3​ in the mixture is respectively:

50%,50%

60%,40%

40%,60%

30%,70%

Q. 3.68 of a mixture of $C a C O_{3}$ and $M g C O_{3}$ is heated to liberate 0.04 mole of $C O_{2}$ . The mole $%$ of $C a C O_{3}$ and $M g C O_{3}$ in the mixture is respectively:

$50%, 50%$

$60%, 40%$

$40%, 60%$

$30%, 70%$