Example Limiting Reactant Calculation:

A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped? How much nitrogen monoxide will be produced in this reaction? 1. First, we need to create a balanced equation for the reaction:

4 NH

3

+ 5 O

2

→ 4 NO + 6 H

2

O

2. Next we can use stoichiometry to calculate how much of one of the reactants will be used in the reaction. Do this by choosing one of the givens and solving for the other reactant. NOTE: It does not matter which reactant you start with. 2.00 g NH 3 1 mol NH 3 17.0 g NH 3 5 mol O 2 32.0 g O 2 4 mol NH 3 1 mol O 2 = 4.71 g O 2 are needed for this reaction Because we only have 4.00 g of oxygen but we need 4.71 g oxygen, we know that oxygen is limiting this reaction.

Oxygen is the limiting reactant.

3. Next we can calculate how much product will be formed. Use stoichiometry to calculate how much product will be formed starting with the limiting reactant as your given. The limiting reactant controls how much product can be formed. 4.00 g O 2 1 mol O 2 32.0 g O 2 4 mol NO 30.0 g NO = 3.00 g NO will be formed 5 mol O 2 1 mol NO 4. Finally, to find the amount of excess reactant, we must calculate how much of the excess reactant (NH 3 ) actually reacted with the limiting reactant (oxygen).

4.00 g O 2 1 mol O 2 32.0 g O 2 4 mol NH 3 17.0 g NH 3 5 mol O 2 1 mol NH 3 = 1.70 g NH 3 We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.

Started with used up have left excess NH 3 = 0.30 g

2.00 g NH 3 - 1.70 g NH 3 = 0.30 g NH 3

Extra Practice Problems a. Which of the reagents is the limiting reagent?

b. What is the amount of each product that will be formed?

c. How much of the excess reagent is left over after the reaction is complete?

1. Consider the following reaction: 3 NH 4 NO 3 + Na 3 PO 4 → (NH 4 ) 3 PO 4 + 3 NaNO 3 Answer the questions above, assuming we started with 30.0 grams of ammonium nitrate and 50.0 grams of sodium phosphate.

Answer: a.

c.

Limiting reactant = ammonium nitrate b. 18.6 grams of ammonium phosphate, 31.9 grams of sodium nitrate 29.5 grams of sodium phosphate left over

2. Consider the following reaction: 3 CaCO 3 + 2 FePO 4 → Ca 3 (PO4) 2 + Fe 2 (CO 3 ) 3 Answer the questions at the top of this sheet, assuming we start with 100. grams of calcium carbonate and 45.0 grams of iron (III) phosphate.

Answer: a. Limiting reactant = iron (III) phosphate

b. 46.3 grams of calcium phosphate, 43.8 grams of iron (III) carbonate c. 54.0 grams of calcium carbonate left over