For the reaction: Pb(NO3)2(aq) + 2KI(aq) --> PbI2(s) + 2KNO3(aq)

Question

For the reaction: Pb(NO3)2(aq) + 2KI(aq) --> PbI2(s) + 2KNO3(aq)

How much non-limiting reagent is still left if you should have made 1.1 grams of PbI2 and you used 1.0 grams of Pb(NO3)2 (your non-limiting reagent)

Answer 1

Pb(NO3)2(aq) + 2KI(aq) --> PbI2(s) + 2KNO3(aq)

1. use stoichiometry:

(1.1g PbI2)*(1 mol PbI2/ 461 g/mol PbI2)*(1 mol Pb(NO3)2/1 mol PbI2)*(331.2 g/mol Pb(NO3)2/ 1 molPb(NO3)2) = .790 g Pb(NO3)2 actually used

2.subtract the given non-limting reagent for the one just solved for:

1.0 - .790= .210 grams Pb(NO3)2 left