Caution: this stuff is difficult to follow at first. 2/23/2019 Limiting Reagents Caution: this stuff is difficult to follow at first. Be patient.

Balloon & Flask Demonstration g of NaHCO3 mL of 3M HCl 1 10 25 2 50 3 100 How can we prove that our conclusions about limiting reagents is correct?

Limiting reagent defined Given: 4NH3 + 5O2  6H2O + 4NO Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2? 4 mol NH3, 5 mol O2 4 mol NH3, 20 mol O2 8 mol NH3, 20 mol O2 4 mol NO, works out exactly 4 mol NO, with leftover O2 8 mol NO, with leftover O2 Here, NH3 limits the production of NO; if there was more NH3, more NO would be produced Thus, NH3 is called the “limiting reagent” 4 mol NH3, 2.5 mol O2 2 mol NO, leftover NH3 In limiting reagent questions we use the limiting reagent as the “given quantity” and ignore the reagent that is in excess …

Limiting reagents in stoichiometry 4NH3 + 5O2  6H2O + 4NO E.g. How many grams of NO are produced if 4 moles NH3 are burned in 20 mol O2? Since NH3 is the limiting reagent we will use this as our “given quantity” in the calculation # g NO= 4 mol NO 4 mol NH3 x 30.0 g NO 1 mol NO x 4 mol NH3 = 120 g NO 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 …

Solving Limiting reagents 1: g to mol 4NH3 + 5O2  6H2O + 4NO 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 1 mol NH3 17.0 g NH3 x 1.176 mol NH3 = # mol NH3= 20 g NH3 1 mol O2 32.0 g O2 x 0.9375 mol O2 = # mol O2= 30 g O2 A – Once the number of moles of each is calculated we can determine the limiting reagent via a chart …

2: Comparison chart NH3 O2 What we have What we need 1.176 0.937 1.176/0.937 = 1.25 mol 0.937/0.937 = 1 mol 4 5 4/5 = 0.8 mol 5/5 = 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 A - There is more NH3 (what we have) than needed (what we need). Thus NH3 is in excess, and O2 is the limiting reagent.

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 NH3 is burned in 30 g O2? 4NH3 + 5O2  6H2O + 4NO # g NO= 1 mol O2 32.0 g O2 x 4 mol NO 5 mol O2 x 30.0 g NO 1 mol NO x 30 g O2 22.5 g 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= 1 mol NH3 17.0 g NH3 x 4 mol NO 4 mol NH3 x 30.0 g NO 1 mol NO x 20 g NH3 35.3 g NO = 1 mol O2 32.0 g O2 x 4 mol NO 5 mol O2 x 30.0 g NO 1 mol NO x 30 g O2 22.5 g NO =

Practice questions 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?

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

Question 1: shortcut 2Al + 6HCl  2AlCl3 + 3H2 If 25 g aluminum was added to 90 g HCl, what mass of H2 will be produced? 1 mol Al 27.0 g Al x 3 mol H2 2 mol Al x 2.0 g H2 1 mol H2 x # g H2= 25 g Al = 2.78 g H2 1 mol HCl 36.5 g HCl x 3 mol H2 6 mol HCl x 2.0 g H2 1 mol H2 x # g H2 = 90 g HCl = 2.47 g H2

N2 is the limiting reagent 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= 1 mol N2 28.0 g N2 x 2 mol NH3 1 mol N2 x 17.0 g NH3 1 mol NH3 x 20 g N2 = 24.3 g H2 # g NH3 = 1 mol H2 2.0 g H2 x 2 mol NH3 3 mol H2 x 17.0 g NH3 1 mol NH3 x 5.0 g H2 = 28.3 g H2 N2 is the limiting reagent

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= 1 mol Al 27.0 g Al x 2 mol Al2O3 4 mol Al x 102.0 g Al2O3 1 mol H2 x 10.0 g Al = 18.9 g Al2O3 # g Al2O3= 1 mol O2 32.0 g O2 x 2 mol Al2O3 3 mol O2 x 102.0 g Al2O3 1 mol H2 x 20.0 g O2 = 42.5 g Al2O3

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= 1 mol C3H8 44.0 g C3H8 x 3 mol CO2 1 mol C3H8 x 44.0 g CO2 1 mol CO2 x 15.0 g C3H8 = 45.0 g CO2 # g CO2= 1 mol O2 32.0 g O2 x 3 mol CO2 5 mol O2 x 44.0 g CO2 1 mol CO2 x 60.0 g O2 = 49.5 g CO2 5. Limiting reagent questions give values for two or more reagents (not just one)

Question 2 1 mol N2 28 g N2 x # mol N2= 20 g N2 0.714 mol N2 = 1 mol H2 2 g H2 x # mol H2= 5.0 g H2 2.5 mol H2 = N2 H2 What we have What we need 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.

3 4Al + 3O2  2 Al2O3 1 mol Al 27 g Al x 0.37 mol Al = # mol Al = 1 mol O2 32 g O2 x 0.625 mol O2 = # mol O2 = 20 g O2 There is more than enough O2; Al is limiting Al O2 0.37 mol 0.625 mol What we have What we need 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 2 mol Al2O3 4 mol Al x 102 g Al2O3 1 mol Al2O3 x # g Al2O3 = 0.37 mol Al 18.87 g Al2O3 =

4 C3H8 + 5O2  3CO2 + 4H2O 1 mol C3H8 44 g C3H8 x 0.34 mol C3H8 = 1 mol O2 32 g O2 x 1.875 mol O2 = # mol O2 = 60 g O2 We have more than enough O2, C3H8 is limiting C3H8 O2 What we have 0.34 mol 1.875 mol 0.34/.34 = 1 mol 1.875/0.34 = 5.5 mol Need 1 mol 5 mol # g CO2 = 44 g CO2 1 mol CO2 x 3 mol CO2 1 mol C3H8 x 0.34 mol C3H8 45 g CO2 =

Limiting Reagents: shortcut MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced? # g AgCl= 1 mol MgCl2 95.21 g MgCl2 x 2 mol AgCl 1 mol MgCl2 x 143.3 g AgCl 1 mol AgCl x 25 g MgCl2 75.25 g AgCl = # g AgCl= 1 mol AgNO3 169.88 g AgNO3 x 2 mol AgCl 2 mol AgNO3 x 143.3 g AgCl 1 mol AgCl x 68 g AgNO3 57.36 g AgCl = For more lessons, visit www.chalkbored.com