1. Solving Limiting Reactant Problems

2. Background In limiting reactant problems, we have the amounts (masses or mols) of two of the reactants. The problem is always to find out which one of the reactants is “limiting.” The limiting reactant is the one that we will run out of first. The other reactant will be in excess and will not be totally used up in the reaction.

3. Background We will be using the same techniques we use to solve ideal stoichiometric problems. We will use molar masses and mole ratios to determine the limiting reactant. We will also use these values to figure out how much product we can expect from that limiting reactant.

4. Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l)

5. Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) First, we list the molar masses of the compounds.

6. Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 First, we list the molar masses of the compounds.

7. Next, put in our known values and our unknown quantity. Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0

8. Next, put in our known values and our unknown quantity. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0

9. Next, calculate the number of moles of the knowns. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0

10. Next, calculate the number of moles of the knowns. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol)

11. n C5H12 = = = 2.15 mol m C5H12 155 g M C5H12 72.0 g/mol m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol)

12. n C5H12 = = = 2.15 mol m C5H12 155 g M C5H12 72.0 g/mol 2.15 m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol)

13. n O2 = = = 10.94 mol m O2 350 g M O2 32.0 g/mol m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.15

14. n O2 = = = 10.94 mol m O2 350 g M O2 32.0 g/mol 10.94 m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.15

15. If we used up all 2.15 mols of the C 5 H 12, we would need 8×2.15 mol = 17.20 mols of O 2. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.94

16. We only have 10.94 mols of O 2. Therefore, O 2 is the limiting reactant. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.94

17. Now, we can calculate the number of mols of CO 2 produced. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.94

18. n CO2 = n O2 × = 10.94 mol × = 6.84 mol coefficient CO 2 5 coefficient O 2 8 coefficient O 2 8 m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.94

19. n CO2 = n O2 × = 10.94 mol × = 6.84 mol coefficient CO 2 5 coefficient O 2 8 coefficient O 2 8 6.84 m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.94

20. Finally, we can calculate the amount of CO 2 produced. m (g) 155 350 ( ? ) Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.946.84

21. m (g) 155 350 ( ? ) m CO2 = n CO2 × M CO2 = (6.84 mol)(44.0 g/mol) = 301 g Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.946.84

22. 301 m (g) 155 350 m CO2 = n CO2 × M CO2 = (6.84 mol)(44.0 g/mol) = 301 g Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.946.84

23. 301 m (g) 155 350 m CO2 = n CO2 × M CO2 = (6.84 mol)(44.0 g/mol) = 301 g Example How much carbon dioxide is produced in the combustion reaction of 155 g of pentane, C 5 H 12, with 350 g of oxygen gas, O 2 ? C 5 H 12 (l) + 8O 2 (g) → 5CO 2 (g) + 6H 2 O(l) M (g/mol) 72.0 32.0 44.0 18.0 n (mol) 2.1510.946.84