2. Limiting Reagents Caution: this stuff is difficult to follow at first. Be patient.

3. g of NaHCO 3 mL of 3M HCl 11025 21050 310100 How can we prove that our conclusions about limiting reagents is correct? Balloon & Flask Demonstration

4. Limiting reagent defined Q - How many moles of NO are produced if __ mol NH 3 are burned in __ mol O 2 ? 4 mol NH 3, 5 mol O 2 4 mol NH 3, 20 mol O 2 8 mol NH 3, 20 mol O 2 Given: 4NH 3 + 5O 2 6H 2 O + 4NO 4 mol NO, works out exactly 4 mol NO, with leftover O 2 8 mol NO, with leftover O 2 Here, NH 3 limits the production of NO; if there was more NH 3, more NO would be produced Thus, NH 3 is called the limiting reagent 4 mol NH 3, 2.5 mol O 2 In limiting reagent questions we use the limiting reagent as the given quantity and ignore the reagent that is in excess … 2 mol NO, leftover NH 3

5. Limiting reagents in stoichiometry E.g. How many grams of NO are produced if 4 moles NH 3 are burned in 20 mol O 2 ? Since NH 3 is the limiting reagent we will use this as our given quantity in the calculation 4NH 3 + 5O 2 6H 2 O + 4NO 4 mol NO 4 mol NH 3 x # g NO= 4 mol NH 3 = 120 g NO 30.0 g NO 1 mol NO x 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 …

6. Solving Limiting reagents 1: g to mol 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 = A – Once the number of moles of each is calculated we can determine the limiting reagent via a chart …

7. 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 ** You should have 1 mol in the same column twice in order to make a comparison 4 5 4/5 = 0.8 mol 5/5 = 1 mol 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. 2: Comparison chart

8. 3: Stoichiometry (given = limiting) So far we have followed two steps … 1) Expressed all chemical quantities as moles 2) Determined the limiting reagent via a chart Finally we need to … 3) Perform the stoichiometry using the limiting reagent as the given quantity 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 # g NO= 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

9. 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

10. 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?

11. 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

12. Question 1: shortcut 2Al + 6HCl 2AlCl 3 + 3H 2 If 25 g aluminum was added to 90 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

13. 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

14. 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

15. 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)

16. 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 =

17. 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 18.87 g Al 2 O 3 = 102 g Al 2 O 3 1 mol Al 2 O 3 x

18. 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 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

19. Limiting Reagents: shortcut MgCl 2 + 2AgNO 3 Mg(NO 3 ) 2 + 2AgCl If 25 g magnesium chloride was added to 68 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 For more lessons, visit www.chalkbored.com www.chalkbored.com