CBSE Class 11-science Answered
a sample of CaCo3 is 50% pure on heating 1.12L of Co2(al STP) is obtained.residue left is- a)7.8 b)3.8 c)2.8 d)8.9
Asked by | 02 Dec, 2012, 08:28: PM
Expert Answer
CaCO3 decomposes as following reaction:
CaCO3 ? CaO + CO2
According to the reaction, 1 mol CaCO3 produces 1 mol CO2
At STP 1mol CO2 = 22.4L
We have 1.12L = 1.12/22.4 = 0.05 mol CO2
Therefore we had 0.05 mol CaCO3 to start.
Molar mass CaCO3 = 100g/mol
Now, we had 0.05mol = 0.05*100 = 5g CaCO3
we started with 5g CaCO3 and 5g Impurity
Molar mass CaO = 56g/mol
Mass of CaO remaining = 5* 56/100 = 2.8g
Total mass residue = 5+2.8 = 7.8g
Choice 1 is correct.
We have 1.12L = 1.12/22.4 = 0.05 mol CO2
Therefore we had 0.05 mol CaCO3 to start.
Molar mass CaCO3 = 100g/mol
Now, we had 0.05mol = 0.05*100 = 5g CaCO3
we started with 5g CaCO3 and 5g Impurity
Molar mass CaO = 56g/mol
Mass of CaO remaining = 5* 56/100 = 2.8g
Total mass residue = 5+2.8 = 7.8g
Choice 1 is correct.
Answered by | 14 Dec, 2012, 07:18: PM
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