Question

How do you determine the limiting reagent by using the stoichiometric ratio to determine the limiting reactant?

CuSO4 * 5H2O (s) + 4NH3 (aq) → [Cu(NH3)4]SO4 * H2O (aq) + H2O (l)

Answer 1

One way is to use the "moles of reaction" method.

Explanation:

A mole of reaction is a reaction that uses the stoichiometric amounts of each reactant and product.

For example, for the reaction

`"CuSO"_4"·5H"_2"O" + "4NH"_3 → "Cu"("NH"_3)_4"SO"_4"·H"_2"O" + "H"_2"O"`

1 mol of reaction involves

`"1 mol CuSO"_4"·5H"_2"O, 4 mol NH"_3, "1 mol Cu"("NH"_3)_4"SO"_4"·H"_2"O"`, and `"1 mol H"_2"O"`.

Here's how you can use the concept in a limiting reactant problem.

Identify the limiting reactant

What is the limiting reactant in the reaction between `"20.0 g CuSO"_4"·5H"_2"O"` and `"7.00 g NH"_3`?

We can summarize the information.

`M_text(r):color(white)(mmmm)249.68color(white)(mmll)17.03color(white)(mmmmm)245.75`
`color(white)(mmmm)"CuSO"_4"·5H"_2"O" + "4NH"_3 → "Cu"("NH"_3)_4"SO"_4"·H"_2"O" + "H"_2"O"`
`m"/g":color(white)(mmmm)25.0color(white)(mmmll)8.00`
`n"/mol":color(white)(mmll)0.100color(white)(mmml)0.470`
`"Divide by": color(white)(mm)1color(white)(mmmmml)4`
`"Moles rxn":color(white)(m)0.100color(white)(mmml)0.117`

To get the moles of reaction, we divide the moles of each reactant by its coefficient (i.e. we use the molar ratio of the reactants) in the balanced equation.

I did that for you in the above table.

We see that the copper(II) sulfate is the limiting reactant because it gives fewer moles of reaction.

Calculate the theoretical yield of product

`"Theoretical yield" = 0.100 color(red)(cancel(color(black)("mol rxn"))) × (1 color(red)(cancel(color(black)("mol product"))))/(1 color(red)(cancel(color(black)("mol rxn")))) × "245.75 g product"/(1 color(red)(cancel(color(black)("mol product")))) = "24.6 g product"`

I think this method involves fewer calculations than the "traditional" method of identifying the limiting reactant by calculating the theoretical yield from each reactant.