Friday, April 21, 2006


Here is the solution to the problem.

You need to find the molar masses of each of the reactants to complete it.

You must balance the equation so the mole ratios are correct

PbSO4 + 2Li à Li2SO4 + Pb

Now you have the mole ratios

Limiting reactant:

It takes 1 mol of lead sulfate to react with 2 mol lithium, so how many moles are present?

There are 303.2 g of lead sulfate in every mol of lead sulfate (add all the molar masses)

There are 6.941 g of lithium in every mol of lithium.

325g lead sulfate (mol/303.2 g) = 1.07 mol lead sulfate

190 g lithium (mol/6.941g) = 27.38 mol lithium

In the balanced equation, for every one mol of lead sulfate, you need 2 moles lithium, you are given way more that 2 moles of lithium, so lead sulfate is the limiting reactant, so it is the one you use in stoichiometry

1.07 mol lead sulfate(1 mol lithium sulfate/1 mol lead sulfate)(104.9g lithium sulfate/mol lithium sulfate) = 112.29 g lithium sulfate

1.07 mol lead sulfate(1 mol lead/1mol lead sulfate)(207.2g lead/mol lead sulfate) = 221.7 g lead

There you have it. Not so bad, right?


Adam Cates said...

where is the solution to me not caring?

Coach Cates said...

You veddy funny man!