1) 800 mL of 0.025 M Pb(NO3)2 is mixed with 500 mL of 0.030 M K2CrO4. Calculate the grams of PbCrO4 that form.

Question

I am confused on how to do this I got the answer it should be 4.85 g PbCrO4, but I would want to know how to get that. And also Limiting reactant is K2CrO4, but I don't know how to get limiting reactant from this kind of problem? Please show step by step how to do this! Thank you.

Answer 1

First set up the balanced equation:

K2CrO4 + Pb(NO3)2 -> PbCrO4 + 2KNO3

800ml of .025M solution is 0.02 moles
500ml of .030M solution is 0.015 moles

So, since you have fewer moles of K2CrO4, that will probably be the limiting reagent.

In fact, since 1 mole of the reactants produces 1 mole of PbCrO4, you will wind up with only .015 moles, having used up all the K2CrO4.

.015 moles of PbCrO4 = .015*(207.2+52+4*16) = 4.848 g

Answer 2

Thanks; I really appreciate it!