Calculate the percentage composition of the elements present in lead nitrate. How many Kg of O_(2) can be obtained from 50 kg of 70% pure lead nitrate?

Calculate the percentage composition of the elements present in lead n

1.	Calculate the percentage composition of the elements present in lead nitrate. How many Kg of O_(2) can be obtained from 50 kg of 70% pure lead nitrate?
Answer» Solution :Lead nitrate = `Pb(NO_(3))_(2)` Molecular mass of lead nitrate = 207 + (14`xx`2)+ (16`xx` 6) = 207 + 28 + 96 = 331 G / mol. 331 g of lead nitrate contains 96 g of oxygen. `:.` `50xx10^(3)` g of lead nitrate will contain `96/331 xx 50 xx 10^(3)` = 14501.5 g = 14.501 Kg of oxygen. 100 % PURE lead nitrate contains 14.501 Kg of oxygen. 70 % pure lead nitrate will contain `14.501/100 xx 70` = 10.15 Kg of oxygen. `:.` 70 % pure lead nitrate will contain 10.15 Kg of oxygen.