Question

How many grams of lead chromate are formed from the reaction of 15.0ml of 0.40m potassium chromate with 15 ml of 0.15 m of lead nitrate?

Answer 1

the number grams of lead chromate formed is calculated as follows

write the reaction equation
Pb(No3)2 + K2CrO4 ----> 2KNO3 + PbCrO4

then calculate moles of each reagent

that is moles of Pb(No3)2 = 15 x0.40/1000= 6 x10^-3 moles
moles of K2CrO4 = 15 x0.15/1000= 2.25 x10^-3moles

K2CrO4 moles are the limiting reagent therefore by use of reacting ratio between K2CrO4 to PbcrO4 the moles of pbcro4 is
2.25 x10^-3 moles

mass is therefore = (2.25 x10^-3) x 323.19(molar mass of PbCrO4) = 0.727 g