Question

What is the mass of PbI2, that will precipitate if 10.2 g of Pb(NO3)2, is mixed with 5.73 g of Kl in a sufficient quantity of H2O? (a) 2.06 g (b) 4.13 g (c)7.96 g (d) 15.9g

Answer 1

answer : option (c) 7.96g

explanation : reaction between lead nitrate and potassium iodide is given as... Pb(NO3)2 + 2KI → PbI2(↓) + 2KNO3

here it is clear that , 1 mole of Pb(NO3)2 reacts with 2 moles of KI.

we know, molar mass of Pb(NO3)2 = 331.2 g/mol

and molar mass of KI = 166 g/mol

so, 331.2 g of Pb(NO3)2 reacts with 332 g of KI.

or, 10.2g of Pb(NO3)2 will react with 332 × 10.2/331.2 = 10.224 g of KI

but here given 5.73 of KI

so, KI is limiting reagent

now, 2mole of KI produces 1 mole of PbI2

or, 332 of KI produces 461g/mol of PbI2.

or, 5.73 g of Pb(NO3)2 produces 461/332 × 5.73 ≈ 7.96g

hence, option (c) is correct.

Answer 2

Given : 10.2 g of Pb(NO3)2,

5.73 g of KI .

Find : Mass of PbI2 which will be precipitated.

Solution :

• So , firstly for finding the mass of PbI2 which will be precipitated when 10.2 g of Pb(NO3)2, is mixed with 5.73 g of Kl in a sufficient quantity of H2O , we should know about the limiting reagent.

• Limiting reagent is defined as the reactant in a chemical reaction which is totally consumed during the chemical reaction.

• So, here for it the equation is :

Pb(NO3)2 + 2KI —> PbI2 + KNO3

331.2 g. 2*166 g. 461 g

10.2g. 5.73g. ?

• The molecular mass of Pb(NO3)2 is 331.2 g and for KI it is 166 g ,as 2 molecules of KI present so it will be 2*166 g . And the PI2 molecular mass is 461 g

• So, from here we can see KI is limiting reagent as it will completely consumed in the reaction.

• So, KI will decide the product.

• So, 5.73g of KI will gives = 461/332 *5.73 g = 7.96g.

• So, the correct option is c)7.96 g , means the mass of PbI2 precipitated in the given condition is 7.96g