chem question about limiting reactants?

Question

In the reaction
2AgI + HgI2= Ag2HgI4
5.59 g of AgI and 2.63 g of HgI2 were used. Assuming complete reaction of the limiting reagent, what mass of the excess reagent remains?

Answer 1

Molar mass AgI = 234.77 g/mol
Moles AgI = 5.59 / 234.77 = 0.0238

Molar mass HgI2 = 454.399 g/mol
Moles HgI2 = 2.63 / 454.399 = 0.00579

The ratio between AgI and HgI2 is 2 : 1 so HgI2 is is the limiting reactant
2 : 1 = x : 0.00579
x = 0.0116 moles AgI needed for the reaction
Moles AgI in excess = 0.0238 - 0.0116 = 0.0122
0.0122 mole x 234.77 g / mol = 2.86 g of AgI in excess

Answer 2

You'll basically have to do two different but similar calculations to get to the answer. First, start with the mass of AgI and convert that to moles. From the stoichiometry of your equation, you see that you get 1 mole of product for every two moles of AgI. Dividing the moles of AgI by 2 gives you the theoretical moles of Ag2HGI4.

Repeat that calculation starting with the mass of HgI2. Convert the mass into moles, and calculate moles of product that you can form.

Which ever reactant allows you to form the few moles of product is your limiting reactant.

From the moles of limiting reactant, calculate the number of moles of the excess reactant that were used. Convert that number of moles to grams, and subtract that mass from the original mass of your excess reactant.

Hope this helps...