1.

A mixture of CaCl2 and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all Ca2+ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g CaO. (Atomic mass of Ca = 40). Following reactions take place:CaCl2 + Na2CO3 → CaCO3 + 2NaClCaCO3 → CaO + CO2The number of mol of CaCl2 in the mixture isOptions:0.56 molWrong 1.11 molShould have chosen 0.01 mol3.33 mol

Answer» A mixture of CaCl2 and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all Ca2+ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g CaO. (Atomic mass of Ca = 40). Following reactions take place:

CaCl2 + Na2CO3 → CaCO3 + 2NaCl

CaCO3 → CaO + CO2

The number of mol of CaCl2 in the mixture is

Options:

0.56 mol

Wrong
1.11 mol

Should have chosen
0.01 mol

3.33 mol


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