How many grams of lead product would theoretically be produced between 17.0 g potassium iodide, KI, and 25.0 g of lead (II) nitrate, Pb(NO3)2?

Explanation:

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In this case, since the undergoing chemical reaction is:

Thus, given the masses of potassium iodide (166.0028  g/mol) and lead (II) nitrate (331.2 g/mol), we compute the yielded moles of lead (II) iodide in order to identify the limiting reactant:

Thus, since potassium iodide yields the fewest moles of product, we infer that 0.0512 moles of lead (II) iodide are theoretically yielded, so the associated mass is:

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