If 10.4 grams of iron metal react 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 Fe(NO3)3 + Ag

71.1

Explanation:

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

sure what is it? ^w^

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