A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is

Question

A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is (A) 15.2% (B) 32.1% (C) 21.8% (D) 11.07%

Tardigrade Admin · Accepted Answer

Moles =( text Weight in gm / text Molecular weight ) Moles of CaO =(1.62/56) Moles of C a C l2=(1.62/56) Mass of CaCl 2=(1.62/56) × 111=3.21 gm Mass % of C a C l2=(3.21/10) × 100=32.1 %

Q. A 10.0g sample of a mixture of calcium chloride and sodium chloride is treated with Na2​CO3​ to precipitate the calcium as calcium carbonate. This CaCO3​ is heated to convert all the calcium to CaO and the final mass of CaO is 1.62gms. The % by mass of CaCl2​ in the original mixture is

15.2%

32.1%

21.8%

11.07%

Moles = Molecular weight Weight in gm ​ Moles of CaO=561.62​ Moles of CaCl2​=561.62​ Mass of CaCl2​=561.62​×111=3.21gm Mass % of CaCl2​=103.21​×100=32.1%

Moles $= \frac{Weight in gm}{Molecular weight}$
Moles of $C a O = \frac{1.62}{56}$
Moles of $C a C l_{2} = \frac{1.62}{56}$
Mass of $C a C l_{2} = \frac{1.62}{56} \times 111 = 3.21 g m$
Mass $%$ of $C a C l_{2} = \frac{3.21}{10} \times 100 = 32.1%$