Question

Calculate the percentage composition of the elements present in lead nitrate. How many Kg of O₂ can be obtained from 50 kg of 70% pure lead nitrate?

Answer 1

Lead nitrate = Pb (NO₃)₂ 

Molecular mass of lead nitrate = 207 + (14 x 2) + (16 x 6) 

= 207 + 28 + 96 

= 331 g/mol. 

331 g of lead nitrate contains 96 g of oxygen. 

∴ 50 x 10³ g of lead nitrate will contain 96/331 x 50 x 10³ 

= 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 x 70 = 10.15 Kg of oxygen. 

\(\therefore\)  70 % pure lead nitrate will contain 10.15 Kg of oxygen.