# 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)...

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?

jeew-m | College Teacher | (Level 1) Educator Emeritus

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`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`

Mole ratio

`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.