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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.

  • chemistry -

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

  • chemistry -

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

  • chemistry -

    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.

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