How To Find the Limiting Reactant – Limiting Reactant Example


Ammonia Ball and Stick Model

3D ball and stick model of the ammonia molecule. Todd Helmenstine

Many chemical reactions take place until one of the reactants run out. This reactant is known as the limiting reactant. Often it is straightforward to determine which reactant will be the limiting reactant, but sometimes it takes a few extra steps.

For example, burning propane in a grill. The propane and oxygen in the air combust to create heat and carbon dioxide. You are obviously more likely to run out of propane long before you run out of oxygen in the air. This makes the propane the limiting reactant. Other reactions aren’t quite as easy.

This example problem will show how to use the stoichiometric ratios between the reactants given in the balanced chemical equation to determine the limiting reactant.

Find the Limiting Reactant Example

Question: Ammonia (NH3) is produced when nitrogen gas (N2) is combined with hydrogen gas (H2) by the reaction

N2 + 3 H2 → 2 NH3

50 grams of nitrogen gas and 10 grams of hydrogen gas are reacted together to form ammonia. Which of the two gasses will run out first? (Which gas is the limiting reactant?)

Answer:  The reaction shows us for every mole of N2 consumed, 3 moles of H2 is also consumed. We need 3 moles of hydrogen gas for every mole of nitrogen gas. The first thing we need to find out is the number of moles of each gas is on hand.

N2 Gas: How many moles of nitrogen gas is 50 grams? One mole of nitrogen is 14.007 grams, so one mole of N2 will weigh 28.014 grams.

Limiting Reactant Example Step 1

Limiting Reactant Example Step 2

x moles N2 = 1.78

H2 Gas: How many moles of hydrogen gas is 10 grams? One mole of hydrogen is 1.008 grams so one mole of H2 is 2.016 grams.

Limiting Reactant step 3
Limiting Reactant Example Step 4
x moles H2 = 4.96

Now we know the number moles of each reactant, we can use the ratio from the chemical equation to compare the amounts.  The ratio between hydrogen gas and nitrogen gas should be:

Limiting Reactant Example Step 5

If we divide our moles of H2 into moles of N2, our value will tell us which reactant will come up short. Any value greater than the above ratio means the top reactant is in excess to the lower number. A value less than the ratio means the top reactant is the limiting reactant. The key is to keep the same reactant on top as the step above.

Limiting Reactant Example 6
2.79

Since our value is less than the ideal ratio, the top reactant is the limiting reactant. In our case, the top reactant is the hydrogen.

Answer: Hydrogen gas is the limiting reactant.

It doesn’t matter which reactant you put on top when you do this type of problem as long as you keep it the same throughout the calculations. If we had put nitrogen gas on top instead of hydrogen the ratio would have worked out the same way. The ideal ratio would have been 13 and the calculated ratio would have been 0.358 ( 1.78/4.96 ). The value would have been greater than the ideal ratio so the bottom reactant in the ratio would be the limiting reactant. In this case, it is the hydrogen gas.

 

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