Question

If 10.4 grams of iron metal reacts with 28.4 grams of silver nitrate, how many grams of iron nitrate can be formed and how many grams of the excess reactant will be left over when the reaction is complete? Show all of your work.

unbalanced equation: Fe + AgNO3 “yields”/ Fe(NO3)3 + Ag

Answer 1

first you want to calculate the moles of each reactant and then determine the limiting reactant

Answer 2

Can I just give you what I have first? @amematsro Just let me know if it's right. So

1 mol Fe = 10.4 g/55.85 g/mol = 0.186
1 mol AgNo3 = 28.4 g/169.87 g/mol = 0.178 mol AgNo3

then since Ag:Fe is 1:3, AgNo3 is the limiting reactant

So now

0.178 moles * 1/3 * 241.83 g/mol Fe(NO3)3 = 14.35 g Fe(NO3)3

Excess reactant: 0.178 moes AgNO3 * 1/3 = 0.059

0.186 - 0.059 = 0.127 moles Fe * 55.85 g/mol Fe = 7.1 g Fe excess

Answer 3

yeah that is exactly what you're supposed to do