1.00g of mixture of calcium carbonate and calcium oxide liberated 0.33g of carbon dioxide on strong heating. The percentage of calcium oxide in the mixture is? (Ca = 40, C = 12, O = 16)
5
15
25
35
50
Correct answer is C
CaCO3 + CaO \(\to\) 2CaO + 2CO2
44gm of CO3 is liberated by 100gm of CaCO3
0.33gm of CO2 will be liberated by \(\frac{100}{44}\) x 0.33
= 0.75 of CaCO3
Amount of CaO = 1 - 0.75
= 0.25
% of CaO = \(\frac{0.25}{1} \times \frac{100}{1}\)
= 25%