PRODUCTS

1. PRODUCTS

2. WHY CAN’T I USE ALL MY BUILDING BLOCKS? (REACTANTS) • WE DON’T HAVE ENOUGH OF EACH OF THE BLOCKS TO BUILD FULL PRODUCTS • SOME REACTANTS ARE LEFT • OVER BUT ONE REACTANT • GETS COMPLETELY USED UP!

3. Limiting Reagents STOICHIOMETRY

4. Limiting Reagent • the reactant which is totally consumed when the chemical reaction is complete “the amount of product formed is limited by this reactant”

5. Given: 4NH3 + 5O2 6H2O + 4NO Limiting reagent Q - How many moles of NO are produced if 4 mol NH3 are burned in 5 mol O2? 4 mol NO, works out PERFECTLY – both reactants are completely used up

6. Given: 4NH3 + 5O2 6H2O + 4NO Q - How many moles of NO are produced if 4 mol NH3 are burned in 20 mol O2? Here, NH3 limits the production of NO; If there was more NH3, more NO would be produced NH3 is called the “limiting reagent” and O2 is in “excess 4 mol NO, with leftover O2

7. On your worksheet: How many moles of NO are produced if 4 mol NH3 are burned in 2.5 mol O2? 2 mol NO, with leftover NH3 • Here, O2 limits the production of NO; if there was more O2, more NO would be produced • Thus, O2 is called the “limiting reagent” and NH3 is in excess!

8. Use a comparison chart between what we have and what the balanced equation says we need… How can I tell which reactant is the limiting reagent?

9. Ok… let’s try it!! 4NH3 + 5O2 6H2O + 4NO 3.2 mol NH3 reacts with 1.6 mol O2 -which reactant will limit the production of the reactants? Comparison chart 3.2 1.6 3.2/1.6 = 2 mol 1.6/1.6 = 1 mol 4/5 = 0.8 5/5 = 1

10. There is more NH3 than needed to react all the O2. So O2 is the limiting reagent which makes NH3 the excess reagent! • Now you can use the limiting reagent moles to calculate how much product you can make!

11. 4 mol NO x 4 mol NH3 4NH3 + 5O2 6H2O + 4NO Limiting reagents in stoichiometry How many moles of NO are produced if 0.25 moles NH3 are burned in 0.56 mol O2? (make a chart) 0.25 mol NH3 is the limiting reagent NO mol= 0.25 mol NH3 =0.25 mol NO

12. Complete questions 3 and 4 on the worksheet!

13. Al(s) + MnO2(aq) Al2O3(aq) + Mn(s) 4 3 2 3 3. What is the limiting reactant when 0.1372 mol of aluminum reacts with 0.1264 mol of MnO2? How many moles of the aluminum product should be yielded from this reaction? 0.1372 0.1264 0.1372/0.1264 1.085 0.1264/0.1264 1 4/3 = 1.33 3/3 = 1

14. So, based on the balanced equation, to use all the MnO2 I would need more Aluminum than I have? Aluminum is the LIMITING REACTANT! Mol Al2O3 made = 0.1372 mol Al x 2 mol Al2O3 4 mol Al = 0.0686 mol

15. If 0.434 mol of both reactants are combined, which will be the limiting reagent? What is the theoretical (predicted) maximum mass of manganese that can be yielded from this reaction? LIMITING REACTANT IS aluminum Theoretical maximum of Mn is 0.3255 mol which (using mm of Mn) converts to 17.88 g

16. Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles we must first determine moles, then decide which is limiting …

17. 1 mol O2 1 mol NH3 1.176 mol NH3 0.9375 mol O2 x x = = 32.0 g O2 17.0 g NH3 4NH3 + 5O2 6H2O + 4NO Solving Limiting reagents mass to mole Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2? A - First we need to calculate the number of moles of each reactant # mol NH3= 20 g NH3 # mol O2= 30 g O2

18. A – Once the number of moles of each is calculated-find the LR… Comparison chart 1.176 0.937 1.176/0.937 = 1.25 mol 0.937/0.937 = 1 mol 4 5 *Choose the smallest value to divide each by

19. A - There is more NH3 (what we have) than needed (what we need). Thus NH3 is in excess, and O2 is the limiting reagent.

20. Stoichiometry 1) Expressed all chemical quantities as moles 2) Determined the limiting reagent via a chart 3) Use the limiting reagent to determine how much product can be made

21. = = 35.3 g NO 22.5 g NO 1 mol NH3 1 mol O2 4 mol NO 4 mol NO 30.0 g NO 30.0 g NO x x x x x x 32.0 g O2 17.0 g NH3 4 mol NH3 5 mol O2 1 mol NO 1 mol NO Limiting Reagents: “shortcut” • Limiting reagent problems can be solved another way (without using a chart)… • Do two separate calculations using both given quantities. The smaller answer is correct. Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2? 4NH3+5O26H2O+4NO # g NO= 20 g NH3 30 g O2

22. 2Al + 6HCl  2AlCl3 + 3H2 • If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced (try this two ways – with a chart & using the shortcut)? • N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? • What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? • When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? • How can you tell if a question is a limiting reagent question vs. typical stoichiometry? • If 25 g magnesium chloride was added to 68 g silvernitrate,whatmassofAgClwillbeproduced? • MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl Practice questions

23. 1 mol Al 1 mol HCl 1 mol HCl 3 mol H2 2.0 g H2 x x x x x 27.0g Al 36.5g HCl 36.5 g HCl 6 mol HCl 1 mol H2 #mol Al = 25 g Al = 0.926 mol 1 # mol HCl= 90 g HCl = 2.466 mol 0.926 2.466 HCl is limiting. 2.466/0.926 = 2.7 mol 0.926/0.926 = 1 mol 2 6 2/2 = 1 mol 6/2 = 3 mol # g H2 = 90 g HCl = 2.47 g H2

24. 1 mol HCl 1 mol Al 3 mol H2 3 mol H2 x x x x 2 mol Al 6 mol HCl 36.5 g HCl 27.0 g Al = 2.47 g H2 = 2.78 g H2 2.0 g H2 2.0 g H2 x x 1 mol H2 1 mol H2 Question 1: shortcut 2Al + 6HCl  2AlCl3 + 3H2 If 25.0 g aluminum was added to 90.0 g HCl,whatmassofH2willbeproduced? # g H2= 25 g Al # g H2 = 90 g HCl

25. = 0.714 mol N2 = 2.5 mol H2 1 mol H2 1 mol N2 x x 28 g N2 2 g H2 Question 2 # mol N2= 20 g N2 # mol H2= 5.0 g H2 0.714 mol 2.5 mol 0.714/0.714 = 1 mol 2.5/0.714 = 3.5 mol 1 mol 3 mol We have more H2 than what we need, thus H2 is in excess and N2 is the limiting factor.

26. 1 mol H2 1 mol N2 2 mol NH3 2 mol NH3 x x x x 1 mol N2 3 mol H2 2.0 g H2 28.0 g N2 = 24.3 g H2 = 28.3 g H2 17.0 g NH3 17.0 g NH3 x x 1 mol NH3 1 mol NH3 Question 2: shortcut N2 + 3H2 2NH3 If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? # g NH3= 20 g N2 # g NH3 = 5.0 g H2 N2 is the limiting reagent

27. 1 mol O2 1 mol Al 0.625 mol O2 0.37 mol Al x x = = 32 g O2 27 g Al What we have = 18.9 g Al2O3 What we need 2 mol Al2O3 102 g Al2O3 x x 4 mol Al 1 mol Al2O3 3 # mol Al = 10 g Al # mol O2= 20 g O2 4Al + 3O2 2 Al2O3 There is more than enough O2; Al is limiting 0.37 mol 0.625 mol 0.37/.37 = 1 mol 0.625/0.37 = 1.68 mol 4 mol 3 mol 4/4 = 1 mol 3/4 = 0.75 mol # g Al2O3 = 0.37 mol Al

28. 1 mol O2 1 mol Al 2 mol Al2O3 2 mol Al2O3 x x x x 4 mol Al 3 mol O2 32.0 g O2 27.0 g Al = 42.5 g Al2O3 = 18.9 g Al2O3 102.0 g Al2O3 102.0 g Al2O3 x x 1 mol H2 1 mol H2 Question 3: shortcut 4Al + 3O2 2 Al2O3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? # g Al2O3= 10.0 g Al # g Al2O3= 20.0 g O2

29. 1 mol C3H8 1 mol O2 0.34 mol C3H8 1.875 mol O2 x x = = 32 g O2 44 g C3H8 = 45.0 g CO2 3 mol CO2 44 g CO2 x x 1 mol C3H8 1 mol CO2 4 # mol C3H8 = 15 g C3H8 # mol O2= 60 g O2 C3H8 + 5O2 3CO2 + 4H2O We have more than enough O2, C3H8 is limiting 0.34 mol 1.875 mol What we have 0.34/.34 = 1 mol 1.875/0.34 = 5.5 mol Need 1 mol 5 mol # g CO2 = 0.34 mol C3H8

30. 1 mol O2 1 mol C3H8 3 mol CO2 3 mol CO2 x x x x 1 mol C3H8 5 mol O2 32.0 g O2 44.0 g C3H8 = 45.0 g CO2 = 49.5 g CO2 44.0 g CO2 44.0 g CO2 x x 1 mol CO2 1 mol CO2 Question 4: shortcut C3H8 + 5O2 3CO2 + 4H2O When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? # g CO2= 15.0 g C3H8 # g CO2= 60.0 g O2 5. Limiting reagent questions give values for two or more reagents (not just one)

31. 1 mol AgNO3 1 mol MgCl2 2 mol AgCl 2 mol AgCl x x x x 1 mol MgCl2 2 mol AgNO3 169.88 g AgNO3 95.21 g MgCl2 = = 75.25 g AgCl 57.36 g AgCl 143.3 g AgCl 143.3 g AgCl x x 1 mol AgCl 1 mol AgCl Question 6: shortcut MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl If 25.00 g magnesium chloride was added to 68.00 g silvernitrate,whatmassofAgClwillbeproduced? # g AgCl= 25 g MgCl2 # g AgCl= 68 g AgNO3