What is the theoretical yield (in g of precipitate) when 16.3 ml of a 0.628 m solution of iron(iii) chloride is combined with 16.5 ml of a 0.642 m solution of lead(ii) nitrate?

2.78 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to stoichiometry:

3 moles of reacts with 2 moles of

Thus 0.010 moles of   will give= of

is a limiting reagent as it limits the formation of products and is an excess reagent.

As 3 moles of give = 3 moles of

Thus 0.010 moles of will give= of  

mass of

Thus the theoretical yield (in g of precipitate) when 16.3 mL of a 0.628 M solution of iron(III) chloride is combined with 16.5 mL of a 0.642 M solution of lead(II) nitrate is 2.78 grams

i think its 100.

hope this !