**How to use limiting reactant to determine theoretical yield**

Since the limiting reactant is completely used up first during a reaction, it follows that the limiting reactant determines how much of the product you can possibly make. The amount of product you can possibly make based on the amount of the limiting reactant is called the theoretical yield.

To calculate theoretical yield, you must consider the following:

- If you know the mass of the limiting reactant, you must first use the molar mass of the reactant to convert from mass of reactant to moles of reactant. If you know the moles of the reactant, like we do from finding the limiting reactant, you should proceed directly to step 2.
- Use the mole-mole ratio between the limiting reactant and the product in the balanced chemical equation to convert the moles you got in step 1 to the corresponding moles of the product
- Multiply the moles of product from step 2 by its corresponding molar mass to get the theoretical yield

The above steps are summarized in the following diagram:

Now let’s solve the following problem

**Question 1**

Methane reacts with oxygen to produce carbon dioxide and water according to the following balanced chemical equation:

CH_{4}+ 2O_{2} ———> CO_{2}+ 2H_{2}O

If 8.0 g of CH_{4} is the limiting reactant calculate the theoretical yield of CO_{2 }and H_{2}O.

#### Strategy

**Continuing from question 1**

Since we determined the moles of CH_{4} in question 1, it follows that we must use these moles and the mole-mole ratio from the balanced chemical equation to calculate the corresponding moles of CO_{2} and H_{2}O that will react with it.

If we refer to the balanced chemical equation:

1 mol CH_{4} + 2 mol O_{2} ———> 1 mol CO_{2} + 2 mol H_{2}O

You will notice that: 1 mol CH_{4}= 1 mol CO_{2}

It follows that the conversion factors from this relationship are:

1 mol CH_{4}/1 mol CO_{2} and 1 mol CO_{2}/1mol CH_{4}

Therefore, to convert 0.498 mol CH_{4} to mol of CO_{2},

we will do: 0.489 mol CH_{4} x 1 mol CO_{2}/1 mol CH_{4}= 0.498 mol CO_{2}

Since we now know the corresponding moles of CO_{2} to be 0.498 mol CO_{2}, we will multiply this value by the molar mass of CO_{2}. If we do, we will get:

0.498 mol CO_{2} x 44.01 g/mol CO_{2} = 21.9 g CO_{2}, thus, the theoretical yield of CO_{2} is 21.9 g.

Similarly, for H_{2}O, the mole-mole ratio between H_{2}O and CH_{4} is:

1 mol CH_{4}= 2 mol H_{2}O. The corresponding conversion factors are: 1 mol CH_{4}/2 mol H_{2}O and

2 mol H_{2}O/1 mol CH_{4}

Therefore, to covert 0.498 mol CH_{4} to mol H_{2}O, we will do:

0.489 mol CH_{4}x 2 mol H_{2}O/1 mol CH_{4}= 0.978 mol H_{2}O

Since we now know the corresponding moles of H_{2}O to be 0.978 mol H_{2}O, we will then multiply this value by the molar mass of H_{2}O. If we do, we will get:

0.978 mol H_{2}O x 18.01 g/molH_{2}O = 17.6 g H_{2}O, thus, the theoretical yield of H_{2}O is 17.6 g.

**How to calculate percentage yield**

To calculate percentage yield, you must divide the actual yield by the theoretical yield, and then multiply the result by 100%. Mathematically, we can express this relationship as:

percentage yield = actual yield/theoretical x 100%

**Why calculate percentage yield?**

We calculate percentage yield to determine how much of the product is actually produced as a result of the reaction. As you may have thought, rarely do reactions produce 100% of the theoretical yield. This is because;

- unwanted side reactions may have occurred during the reaction
- separating the desired products from the reaction mixture may have been difficult

Now, imagine that if 10.00 g of CO_{2 }was actually produced as a result of equation (1) above, calculate the percentage yield of CO_{2}

**Strategy**

To calculate percentage yield, we will take the actual yield of CO_{2} divided by the theoretical yield, and then multiply the result by 100%. If we do, we will get:

10.00 g CO_{2}/21.9 g CO_{2 } x 100 % = 45.7%, thus, 45.7% CO_{2} was produced as a result of the reaction.

To determine limiting reactant, click here