If 119.5 g of iron is mixed with 279.50 g of copper (II) nitrate, what is the limiting reagent for the reaction?

Copper ii nitrate is the limiting reagent

Explanation:

The first thing to write here is the equation of reaction;

Fe + Cu(NO3)2 ——> Fe(NO3)2 + Cu

When we talk of limiting reagent, we mean the reactant that is totally consumed upon the completion of the chemical reaction

To get the limiting reagent, we can know this by calculating the mass of a specific product formed from each of the masses of the reactants

Now, we can use the amount of copper solid deposited by each of the reactants to know the limiting reagent here.

Since the mole ratio in all cases is 1 to 1, this will be easy

For the iron, the number of moles reacted is mass of iron/ atomic mass of iron

That would be 119.5/56 = 2.134 moles

Now since the mole ratio is 1 to 1, 2.134 moles of copper will be formed

Thus, the mass of copper produced from the iron will be number of moles * atomic mass of copper = 2.134 * 64 = 136.57g

Now from the nitrate, the number of moles is

mass mixed/ molar mass of copper nitrate

molar mass of copper nitrate is 188g/mol

number of moles is thus 279.5/188 = 1.487 moles

Since the mole ratio is 1 to 1, it means that the number of moles of copper produced too is 1.487 moles

The mass of copper produced from this is number of moles of copper * atomic mass of copper

That will be 1.487 * 64 = 95.15g

Now, since the copper nitrate produces less amount of copper solid, it is the the reagent to be consumed first and thus it is our limiting reagent

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