# How can you tell what is a limiting reactant in a problem?

Feb 4, 2017

The reactant with the fewer moles.

#### Explanation:

$\text{HC"l(aq)+"NaOH"(aq) rarr "NaC"l(aq)+"H"_2"O} \left(l\right)$

If we have $10$ ${\text{cm}}^{3}$ of $1$ $\text{M}$ $\text{HC} l$ reacting with $25$ ${\text{cm}}^{3}$ of $1$ $\text{M}$ $\text{NaOH}$, we know that $\text{HC} l$ is the limiting reactant.

We have $0.01$ $\text{mol}$ of $\text{HC} l$ but $0.025$ $\text{mol}$ $\text{NaOH}$. When both react, we will be left with $0.15$ $\text{mol}$ of $\text{NaOH}$, so $\text{HC} l$ is the limiting reactant.

Feb 4, 2017

You can pick which way is easier for you:

1. Calculate the mass yield and see which reactant corresponds to the smaller mass yield.
2. Use the mass of one chosen reactant to calculate the required mols of the other reactant, and separately calculate the available mols of the other reactant from its mass. Check whether the available mols of the other reactant are less than the required mols.

Consider the following reaction:

$2 \text{H"_2(g) + "O"_2(g) -> 2"H"_2"O} \left(g\right)$

Let's say you had $\text{2.00 g}$ of ${\text{H}}_{2}$ and $\text{4.00 g}$ of ${\text{O}}_{2}$. Determine the limiting reactant.

1. COMPARE MASS YIELDS

$2.00 \cancel{\text{g H"_2) xx cancel("1 mol H"_2)/(1.0079 cancel("g H"_2)) xx cancel("2 mol H"_2"O")/(2 cancel("mols H"_2)) xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O}}$

$=$ $\text{35.8 g H"_2"O}$ possible

$4.00 \cancel{\text{g O"_2) xx cancel("1 mol O"_2)/(31.998 cancel("g O"_2)) xx cancel("2 mol H"_2"O")/(1 cancel("mol O"_2)) xx ("18.015 g H"_2"O")/cancel("1 mol H"_2"O}}$

$=$ $\text{4.50 g H"_2"O}$ possible

By far, ${O}_{2} \left(g\right)$ is the limiting reactant, because it gave a smaller yield and thus limits the amount of product you can make.

2. COMPARE AVAILABLE MOLS

This may be faster:

2.00 cancel("g H"_2) xx cancel("1 mol H"_2)/(1.0079 cancel("g H"_2)) xx "1 mol O"_2/(2 cancel("mols H"_2))

$=$ ${\text{0.992 mols O}}_{2}$ required

4.00 cancel("g O"_2) xx "1 mol O"_2/(31.998 cancel("g O"_2))

$=$ ${\text{0.125 mols O}}_{2}$ available

There is less ${O}_{2}$ available than required, so it is the limiting reactant.