1. Limiting Reactants In a chemical reaction you often have two or more reactants and until now we have assumed that there has been enough of each reactant for the reaction to continue to completion (both reactants totally used up). Limiting reactant questions give both amounts of reactants and we must determine which one runs out first (the limiting reactant).

4. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount –In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

5. Limiting Reactants In the example below, the O 2 would be the excess reagent

6. Steps to help you determine the limiting reactant Determine what information is given and what is required to calculate. Convert the given information to moles. Use the mole ratio to determine the limiting reactant and the excess reactant. Once the limiting reactant is determined, use it in the mole ratio calculation to determine the unknown. Convert the moles to the required units.

7. Limiting Reactant Example 1 Zn + 2 HCl  ZnCl 2 + H 2 If 1.21 mol of zinc is added to 2.65 mol of HCl, which reactant is the limiting reactant? Mole Ratio 1 Zn 2 HCl Set up conversion using one piece of information (you chose) and cross multiply 1 Zn = 1.21 Zn 2 HCl ? (1)(?) = (2)(1.21) Answer 2.42 mol HCl

8. 2.42 mol HCl is needed for the reaction. You have 2.65 mol, so HCl is the excess reactant. This means that Zn is the limiting reactant.

9. If you had used the other piece of information to calculate the limiting reactant, this is what you would have: 1 Zn = ? Zn 2 HCl 2.65 (1)(2.65) = (2)(?) Answer 1.33 mol Zn

10. 1.33 mol Zn is needed for the reaction. You only have 1.21 mol, so Zn is the limiting reactant. This means that HCl is the excess reactant.

11. Once the limiting reactant is determined, you would use it to finish the question.

12. Limiting Reactant Example 2 2 Al + 3 CuSO 4  3 Cu + Al 2 (SO4) 3 If 10.45 g Al reacts with 66.55 g CuSO 4, what is the limiting reactant? Convert Al to moles 10.45 g Al x 1 mol 26.981539g 0.3873 mol Al Convert CuSO 4 to moles 66.55 g CuSO 4 x 1 mol 159.610g 0.4170 mol CuSO 4

13. Mole Ratio 2 Al 3 CuSO 4 Conversion and Cross Multiply 2 Al = ? 3 CuSO 4 0.4170 (3)(?) = (2)(0.4170) Answer 0.2780 mol Al You need 0.2780 mol to complete the reaction and you have 0.3873. You have more than enough Al so this is the excess reactant.

14. If you had used the other piece of information to calculate the limiting, this is what you would have: Mole Ratio2 Al 3 CuSO 4 Conversion and Cross Multiply 2 Al = 0.3873 3 CuSO 4 ? (3)(0.3873) = (2)(?) Answer 0.5810 mol CuSO 4

15. You have 0.4170 mol of CuSO 4 and you need 0.5810 mol, so CuSO 4 is the limiting reactant.

16. Now that you know the limiting reactant, calculate how many mole of Cu will be formed Mole Ratio 3 CuSO 4 3 Cu Conversion and Cross Multiply 3 CuSO 4 = 0.4170 3 Cu ? (3)(0.4170) = (3)(?) Answer 0.4170 mol Cu

17. Limiting Reactant Example 3 C 2 H 4 + 3 O 2  2 CO 2 + 2 H 2 O 2.80 mol ethene and 6.30 mol oxygen gas. What is the limiting reagent? How many grams of CO 2 will be produced?

18. One or the Other Mole Ratio 1 C 2 H 4 3 O 2 Conversion and Cross Multiply 1 C 2 H 4 = 2.80 3 O 2 ? (3)(2.80) = (1)(?) Answer 8.40 mol O 2 Mole Ratio 1 C 2 H 4 3 O 2 Conversion and Cross Multiply 1 C 2 H 4 = ? 3 O 2 6.30 (3)(?) = (1)(6.30) Answer 2.10 mol C 2 H 4

19. You have 6.30 mol of O 2 and you need 8.40. Oxygen gas is the limiting reactant. You have 2.80 mol C 2 H 4 and you only need 2.10. Ethene is the excess reactant.

20. Now that you have the limiting reactant… Mole Ratio 3 O 2 2 CO 2 Conversion and Cross Multiply 3 O 2 = 6.30 2 CO 2 ? (2)(6.30) = (3)(?) Answer 4.20 mol CO 2

21. Calculate grams 4.20 mol CO 2 x 44.010 g 1 mol 184.84 g CO 2 will be formed. (185 g)