2. PRODUCTS

3. 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!

4. Limiting Reagents STOICHIOMETRY

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

6. Limiting reagent Q - How many moles of NO are produced if 4 mol NH 3 are burned in 5 mol O 2 ? Given: 4NH 3 + 5O 2  6H 2 O + 4NO 4 mol NO, works out PERFECTLY – both reactants are completely used up

7. Here, NH 3 limits the production of NO; If there was more NH 3, more NO would be produced NH 3 is called the “limiting reagent” and O 2 is in “excess 4 mol NO, with leftover O 2 Given: 4NH 3 + 5O 2  6H 2 O + 4NO Q - How many moles of NO are produced if 4 mol NH 3 are burned in 20 mol O 2 ?

8. On your worksheet: How many moles of NO are produced if 4 mol NH 3 are burned in 2.5 mol O 2 ? 2 mol NO, with leftover NH 3 Here, O 2 limits the production of NO; if there was more O 2, more NO would be produced Thus, O 2 is called the “limiting reagent” and NH 3 is in excess!

9. How can I tell which reactant is the limiting reagent? Use a comparison chart between what we have and what the balanced equation says we need… NH 3 O2O2 What we haveMole given From the question Mole ratio calculated What we needRatio in balanced equation

10. Ok… let’s try it!! 4NH 3 + 5O 2  6H 2 O + 4NO 3.2 mol NH 3 reacts with 1.6 mol O 2 -which reactant will limit the production of the reactants? NH 3 O2O2 What we have What we need Comparison chart 3.2 1.6 3.2/1.6 = 2 mol 1.6/1.6 = 1 mol 4/5 = 0.85/5 = 1

11. There is more NH 3 than needed to react all the O 2. So O 2 is the limiting reagent which makes NH 3 the excess reagent! Now you can use the limiting reagent moles to calculate how much product you can make!

12. Limiting reagents in stoichiometry How many moles of NO are produced if 0.25 moles NH 3 are burned in 0.56 mol O 2 ? (make a chart) 4NH 3 + 5O 2  6H 2 O + 4NO 4 mol NO 4 mol NH 3 x NO mol= 0.25 mol NH 3 =0.25 mol NO 0.25 mol NH 3 is the limiting reagent

13. Complete questions 3 and 4 on the worksheet!

14. Al (s) + MnO 2(aq)  Al 2 O 3(aq) + Mn (s) 3. What is the limiting reactant when 0.1372 mol of aluminum reacts with 0.1264 mol of MnO 2 ? How many moles of the aluminum product should be yielded from this reaction? AlMnO 2 What we have What we need 0.1372 0.1264 0.1372/0.1264 1.085 0.1264/0.1264 1 3 42 3 4/3 = 1.33 3/3 = 1

15. So, based on the balanced equation, to use all the MnO 2 I would need more Aluminum than I have? Aluminum is the LIMITING REACTANT! Mol Al 2 O 3 made = 0.1372 mol Al x 2 mol Al 2 O 3 4 mol Al = 0.0686 mol

16. 4.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

17. 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 …

18. Solving Limiting reagents mass to mole Q - How many g NO are produced if 20 g NH 3 is burned in 30 g O 2 ? A - First we need to calculate the number of moles of each reactant 4NH 3 + 5O 2  6H 2 O + 4NO 1 mol NH 3 17.0 g NH 3 x # mol NH 3 = 20 g NH 3 1.176 mol NH 3 = 1 mol O 2 32.0 g O 2 x # mol O 2 =30 g O 2 0.9375 mol O 2 =

19. NH 3 O2O2 What we have What we need 1.176 0.937 1.176/0.937 = 1.25 mol 0.937/0.937 = 1 mol *Choose the smallest value to divide each by 4 5 Comparison chart A – Once the number of moles of each is calculated-find the LR…

20. A - There is more NH 3 (what we have) than needed (what we need). Thus NH 3 is in excess, and O 2 is the limiting reagent.

21. 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

22. 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 NH 3 is burned in 30 g O 2 ? 4NH 3 + 5O 2  6H 2 O+ 4NO 4 mol NO 5 mol O 2 x 30 g O 2 22.5 g NO= 30.0 g NO 1 mol NO x 1 mol O 2 32.0 g O 2 x 4 mol NO 4 mol NH 3 x # g NO= 20 g NH 3 35.3 g NO= 30.0 g NO 1 mol NO x 1 mol NH 3 17.0 g NH 3 x

23. Practice questions 1.2Al + 6HCl  2AlCl 3 + 3H 2 If 25 g of aluminum was added to 90 g of HCl, what mass of H 2 will be produced (try this two ways – with a chart & using the shortcut)? 2.N 2 + 3H 2  2NH 3 : If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 3.What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O 2 ? 4.When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 5.How can you tell if a question is a limiting reagent question vs. typical stoichiometry? 6.If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced? MgCl 2 + 2AgNO 3  Mg(NO 3 ) 2 + 2AgCl

24. 1 1 mol Al 27.0 g Al x # mol Al =25 g Al = 0.926 mol # mol HCl =90 g HCl 1 mol HCl 36.5 g HCl x = 2.466 mol AlHCl What we have What we need 0.926 2.466 0.926/0.926 = 1 mol 2.466/0.926 = 2.7 mol 2 6 2/2 = 1 mol 6/2 = 3 mol HCl is limiting. 3 mol H 2 6 mol HCl x # g H 2 = 90 g HCl 2.0 g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x = 2.47 g H 2

25. Question 1: shortcut 2Al + 6HCl  2AlCl 3 + 3H 2 If 25.0 g aluminum was added to 90.0 g HCl, what mass of H 2 will be produced? 3 mol H 2 2 mol Al x # g H 2 =25 g Al = 2.78 g H 2 2.0 g H 2 1 mol H 2 x 1 mol Al 27.0 g Al x 3 mol H 2 6 mol HCl x # g H 2 =90 g HCl = 2.47 g H 2 2.0 g H 2 1 mol H 2 x 1 mol HCl 36.5 g HCl x

26. N2N2 H2H2 What we have What we need Question 2 0.714 mol 2.5 mol 0.714/0.714 = 1 mol 2.5/0.714 = 3.5 mol We have more H 2 than what we need, thus H 2 is in excess and N 2 is the limiting factor. 1 mol 3 mol 1 mol N 2 28 g N 2 x # mol N 2 =20 g N 2 0.714 mol N 2 = 1 mol H 2 2 g H 2 x # mol H 2 =5.0 g H 2 2.5 mol H 2 =

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

28. AlO2O23 4Al + 3O 2  2 Al 2 O 3 1 mol Al27 g Al x # mol Al =10 g Al 0.37 mol Al = 1 mol O 2 32 g O 2 x # mol O 2 =20 g O 2 0.625 mol O 2 = 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 mol3/4 = 0.75 mol What we have What we need There is more than enough O 2 ; Al is limiting 2 mol Al 2 O 3 4 mol Al x # g Al 2 O 3 =0.37 mol Al 18.9 g Al 2 O 3 = 102 g Al 2 O 3 1 mol Al 2 O 3 x

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

30. C3H8C3H8 O2O24 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 1 mol C 3 H 8 44 g C 3 H 8 x # mol C 3 H 8 =15 g C 3 H 8 0.34 mol C 3 H 8 = 1 mol O 2 32 g O 2 x # mol O 2 =60 g O 2 1.875 mol O 2 = 0.34 mol 1.875 mol 0.34/.34 = 1 mol 1.875/0.34 = 5.5 mol 1 mol 5 mol What we have Need 3 mol CO 2 1 mol C 3 H 8 x # g CO 2 = 0.34 mol C 3 H 8 45.0 g CO 2 = 44 g CO 2 1 mol CO 2 x We have more than enough O 2, C 3 H 8 is limiting

31. Question 4: shortcut C 3 H 8 + 5O 2  3CO 2 + 4H 2 O When C 3 H 8 burns in oxygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 3 mol CO 2 1 mol C 3 H 8 x # g CO 2 = 15.0 g C 3 H 8 = 45.0 g CO 2 44.0 g CO 2 1 mol CO 2 x 1 mol C 3 H 8 44.0 g C 3 H 8 x 3 mol CO 2 5 mol O 2 x # g CO 2 = 60.0 g O 2 = 49.5 g CO 2 44.0 g CO 2 1 mol CO 2 x 1 mol O 2 32.0 g O 2 x 5. Limiting reagent questions give values for two or more reagents (not just one)

32. Question 6: shortcut MgCl 2 + 2AgNO 3  Mg(NO 3 ) 2 + 2AgCl If 25.00 g magnesium chloride was added to 68.00 g silver nitrate, what mass of AgCl will be produced? 2 mol AgCl 1 mol MgCl 2 x # g AgCl= 25 g MgCl 2 75.25 g AgCl= 143.3 g AgCl 1 mol AgCl x 1 mol MgCl 2 95.21 g MgCl 2 x 2 mol AgCl 2 mol AgNO 3 x # g AgCl= 68 g AgNO 3 57.36 g AgCl = 143.3 g AgCl 1 mol AgCl x 1 mol AgNO 3 169.88 g AgNO 3 x