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?

Explanation

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.