15 Limiting Reactant Problems: And Solutions

A limiting reactant problem is where in the stoichiometric ratio of the reactants is not given.

In this limiting reactant problems what we determine is, there is a reactant (the limiting) which limits the amount of product that can be obtained or produced.

We can solve the limiting reactant problem very easily by following the below steps:

  • First, write a balanced complete reaction.
  • The reactants should be converted to moles.
  • Then divide by coefficient.
  • The reactant obtained with a smaller number is the limiting reactant.

1. 100g of sucrose combusts with 10.0g of oxygen forming carbon dioxide and water. Which is the limiting reactant?

                    Solution

                    Step 1: Obtaining a balanced chemical equation:

                    C12H22O11 + 12 O2 12 CO2 + 11 H2O

                    Step 2: Converting reactants to moles

So, in the above problem O2 is the limiting reactant (because limiting reactant = reactant that produces least ml of product).

 2. Find the limiting reactant when 4.687g of SF4 reacts with 6.281g of I2O5 to produceIF5 and SO2

Solution

Step 1: Obtaining a balanced chemical equation

5SF4 + 2I2O5 → 4IF5 + 5SO2

Step 2: Converting reactants to mole then dividing by coefficient

rxn 2

So, 0.0094 mol (I2O5) is the limiting reactant as it has the lower value as compared to SF4 (0.00867 mol).

3. 4.5 moles of antimony reacts with 5.5 moles of oxygen according to the balanced equation, what is the limiting reactants? (Antimony →1 × 121.8 = 121.8 g/mol, oxygen (O2) → 2 × 16.0 = 32.0g/mol)

Solution

Step 1: Obtaining a balanced chemical equation

4Sb + 302 → Sb4O6

Step 2: Already given in moles so divide by coefficient

rxn 3

So, Sb (1.125 mol) is the limiting reactant as it has the lower value as compared to O2 (1.83 mol).

4. 6.7 moles of iron react with 8.4 moles of oxygen, what is the limiting reactant? (Iron →1 × 55.9 = 55.9g/mol, oxygen → 2 × 16.0 = 32.0g/mol)

Solution

Step 1: Obtaining a balanced chemical equation

4Fe + 3O2 → 2Fe2 O3

Step 2: already given in moles so divide by coefficient

rxn 4

So, Fe (3.35 mol) is the limiting reactant as it has the lower value as compared to O2 (5.6 mol).

5. 13 moles of phosphorus (P4) reacts with 194 grams of water (H2O). What is the limiting reactant? (P4→4 x 31.0 = 124.0g/mol, H2O→ 2 x 1.0 + 1 x 16.0 = 18.0 g/mol.

Solution

Step 1: Obtaining a balanced chemical equation

P4 + 16H2O →4H3PO4 + 10H2

Step 2: already given in moles so divide by coefficient

rxn 5

So, H2O (6.7 mol) is the limiting reactant as it has the lower value as compared to P4 (130 mol).

6. 17 grams of bismuth (III) nitrate [Bi(NO3)3] reacts with 19 grams of hydrogen sulfide (H2S) according to the balanced equation, what is the limiting reactant? (Bismuth (III) nitrate [Bi(NO3)3] →1 × 209.0 + 3 × 14.0 + 3 × 3 × 16.0 = 395.0g/mol)

Solution

Step 1: Obtaining a balanced Chemical equation

2Bi (NO3)3 + 3H2S → BiS3 + 6HNO3

Step 2: Converting reactants to moles then divide y coefficient

rxn 6 1

So, O2 (0.004 mol) is the limiting reactant as it has the lower value as compared to NH3 (0.105 mol).

7. 0.07 moles of ammonia reacts with 0.11 grams of oxygen, what is the limiting reactant? (ammonia × 1 × 4 + 3 × 1 = 17.0g/ mol, oxygen → 2 × 16 =  32g/mol )

Solution

Step 1: Obtaining a balanced chemical equation

4NH3 + 5O2 → 4NO + 6H2O

Step 2: Converting reactants to moles then divide by coefficient.

rxn 7

So, O2 (0.004 mol) is the limiting reactant as it has the lower value as compared to NH3 (0.105 mol).

8. What mass in grams of aluminium hydroxide could be made from 30g Al2S3 and 20g H2O, What is the limiting agent ?

Solution

Step 1: Obtaining a balanced chemical equation.

Al2S3 + 6H2O → 2Al(OH)3 + 3H2S

Step 2: Converting reactants to moles then divide by coefficient.

rxn 8

So, H2O (28.8g) is the limiting reactant as it has the lower value as compared to Al2S3 (31.17g).

Read more about: 10+ Covalent Bond Types Of Elements: Detailed Insights And Facts

9. What mass in grams of P4O6 could be made from 8.75g P4 and 12.50g O2 ?

Solution

Step 1: Obtaining a balanced chemical equation.

P4 + 3O2 → P4O6                                          

Step 2: Converting reactants to moles then divide by coefficient.

rxn 9

So, P4 (15.5g) is the limiting reactant as it has the lower value as compared to O2 (28.63g).

10. What mass in grams of TiCl4 could be made from 25g TiO2 , 10g Carbon and 40g Cl2 ?

Solution

Step 1: Obtaining a balanced chemical equation.

3TiO2 + 4C + 6Cl2 → 3TiCl4 + 2CO2 + 2CO

Step 2: Converting reactants to moles then divide by coefficient.

rxn 10

 So, Cl2 (53.5g) is the limiting reactant as it has the lower value as compared to TiO2 (59.4g) and C (118g).

11. How many grams of water can be produced from 5.55 grams hydrogen and 4.44 grams oxygen ?

Solution

Step 1: Obtaining a balanced chemical equation.

2H2 + O2 → 2H2O

Step 2: Converting reactants to moles then divide by coefficient.

rxn 11

 So, O2 (0.139 mol) is the limiting reactant as it has the lower value as compared to H2 (2.75 mol).

12. If 3.22 moles of Al reacts with 4.96 moles of HBr, what will be the limiting reactant in this limiting reactant problem ?

Solution

Step 1: Obtaining a balanced chemical equation.

2Al + 6HBr → 2AlBr3 + 3H2

Step 2: Converting reactants to moles then divide by coefficient.

rxn 12

 So, HBr (4.99g) is the limiting reactant as it has the lower value as compared to Al (9.737g).

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13. If 21.44 moles of Si reacts with 17.62 moles of N2, what is the limiting reactant ?

Solution

Step 1: Obtaining a balanced chemical equation.

3Si + 2N2 → Si3N4

Step 2: Already in moles of so divide by coefficient.

rxn 13

So, Si (7.146 mol) is the limiting reactant as it has the lower value as compared to N2 (8.81 mol).

14. Methane gas (CH4) reacts with oxygen by combustion. How many grams of methane are needed to produce 25 grams of water?

Solution

Step 1: Obtaining a balanced chemical equation

CH4 + 2O2 → CO2 + 2H2O

Step 2: Converting reactants to moles then divide by coefficient.

rxn 14

Therefore 11.13g of methane is needed to produce 25 grams of water.

15. Consider the reaction NH3 + O2 → NO + H2O. In an experiment 3.25 g of NH3 are allowed to react with 3.50g of O2. What will be the limiting reactant?

Solution

Step 1: Obtaining a balanced chemical equation

4NH3 + 5O2 → 4NO + 6H2O

Step 2: Converting into moles and then dividing by coefficient.

rxn 15 1

So, O2 (0.0219) is the limiting reactant as it has lower value as compared to NH3 (0.0477).

16. If 75 grams of iron (III) chloride reacts with 125 grams of magnesium oxide. What will be the limiting reactant?

Solution

Step 1: Obtaining a balanced chemical equation

2FeCl3 + 3MgO → Fe2O3 + 3MgCl2

Step 2: Converting into moles and then dividing by coefficient.

rxn 16

So, FeCl3 (37.267g) is the limiting reactant as it has lower value as compared to MgO (166.667g).

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