I balanced an equation between lead(II) nitrate and potassium sulfate and ended up with:

Pb(NO3)2 + K2SO4 --> PbSO4 + 2 KNO3

If 0.87 moles of potassium sulfate react with an excess amount of lead(II) nitrate, how many moles of lead(II) sulfate would be formed?

I know I start with 0.87 moles K2SO4, but I am confused on the stoichiometry part to solve the problem.

Here is a worked example that shows exactly how to do the conversion from mols of one material to mols of another.

http://www.jiskha.com/science/chemistry/stoichiometry.html

mm(zn)=65g/mol m(hcl)=36.5g/mol m(h2)=2g/mol

From the equation, 1 mole of potassium sulfate reacts with lead(II) nitrate to form 1 mol of lead(II) sulfate.

Implying that mole ratio of K2SO4:PbSO4=1:1 therefore, number of moles of PbSO4 that formed =1/1 x0.87= 0.87 moles
NB: The moles of K2SO4 that reacted with excess Pb(NO3)2 are the one that determined the number of moles of PbSO4 to be formed.

To solve the stoichiometry problem, you need to use the balanced equation. The balanced equation tells us the mole ratio between the reactants and products. In this case, the mole ratio between potassium sulfate (K2SO4) and lead(II) sulfate (PbSO4) is 1:1.

Given that you have 0.87 moles of potassium sulfate (K2SO4), you can use the mole ratio to determine the number of moles of lead(II) sulfate (PbSO4) that will be formed. Since the mole ratio is 1:1, it means that for every 1 mole of potassium sulfate (K2SO4) that reacts, 1 mole of lead(II) sulfate (PbSO4) will be formed.

Therefore, the number of moles of lead(II) sulfate (PbSO4) formed can be obtained by multiplying the number of moles of potassium sulfate (K2SO4) by the mole ratio of 1.

Moles of PbSO4 = Moles of K2SO4 = 0.87 moles

So, 0.87 moles of lead(II) sulfate (PbSO4) would be formed when 0.87 moles of potassium sulfate (K2SO4) reacts with an excess amount of lead(II) nitrate.