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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.

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