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Tatiana [17]
3 years ago
15

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 g

rams 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
Chemistry
1 answer:
Trava [24]3 years ago
8 0

Answer:

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

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