32.12mL of 10.0 mol/L potassium chromate solution is reacted with 120.0mL of 2.00 mol/L lead (ll) nitrate solution. The balanced equatoin is this :
2KCrO4 + Pb(NO3)2 -> 2KNO3 + Pb(CrO4)2
a) Determine the limiting reactant
b) What is the maximum mass, in grams, of lead (ll) chromate that can be produced?
1 Answer | Add Yours
`2KCrO_4 + Pb(NO_3)_2 rarr 2KNO_3 + Pb(CrO_4)_2`
Amount of `KCrO_4 = 10/1000xx32.12 = 0.3212mol`
Amount of `Pb(NO_3)_2 = 2/1000xx120 = 0.24mol`
Mole ratio (For reaction)
`KCrO_4:Pb(NO_3)_2 = 2:1`
Mole ratio (in the initial solution)
`KCrO_4:Pb(NO_3)_2 = 0.3212:0.24`
For the reaction we need twice the amount of `KCrO_4` than `Pb(NO_3)_2` .
In the solution ratio is less than this.(`KCrO_4` is less than `Pb(NO_3)_2)`
So all the `KCrO_4` will react.
Reacted `KCrO_4` `= 0.3212`
Reacted `Pb(NO_3)_2` `=0.3212/2 = 0.1606`
`Pb(NO_3)_2:Pb(CrO_4)_2 = 1:1`
Amount of `Pb(CrO_4)_2` formed = 0.1606 mol
Molic mass of `Pb(CrO_4)_2 ` = 439g/mol
Weight of Pb(CrO4)2 formed = `0.1606xx439` = 70.5g
So 70.5g of Pb(CrO4)2 is formed.
We’ve answered 318,944 questions. We can answer yours, too.Ask a question