Stoichiometry

Stoichiometry questions (1) Consider : 4NH3 + 5O2  6H2O + 4NO How many moles of H2O are produced if 0.176 mol of O2 are used? How many moles of NO are produced in the reaction if 17 mol of H2O are also produced? 6 mol H2O 5 mol O2 x 0.2112 mol H2O = # mol H2O= 0.176 mol O2 4 mol NO 6 mol H2O x 11.33 mol NO = # mol NO= 17 mol H2O Notice that a correctly balanced equation is essential to get the right answer

Stoichiometry questions (2) Consider : 4NH3 + 5O2  6H2O + 4NO How many grams of H2O are produced if 1.9 mol of NH3 are combined with excess oxygen? How many grams of O2 are required to produce 0.3 mol of H2O? # g H2O= 6 mol H2O 4 mol NH3 x 18.02 g H2O 1 mol H2O x 51.4 g H2O = 1.9 mol NH3 # g O2= 5 mol O2 6 mol H2O x 32 g O2 1 mol O2 x 8 g O2 = 0.3 mol H2O

Stoichiometry questions (3) Consider : 4NH3 + 5O2  6H2O + 4NO How many grams of NO is produced if 12 g of O2 is combined with excess ammonia? # g NO= 1 mol O2 32 g O2 x 4 mol NO 5 mol O2 x 30.01 g NO 1 mol NO x 12 g O2 9.0 g NO =

DO NOW- Balance Al + HCl  AlCl3 + H2 N2 + H2  NH3 Al + O2  Al2O3 C3H8 + O2  CO2 + H2O 2Al + 6HCl  2AlCl3 + 3H2 N2 + 3H2  2NH3 4Al + 3O2  2 Al2O3 C3H8 + 5O2  3CO2 + 4H2O

Limiting Reagents: 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 =

Limiting Reagents 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

Practice questions-LR 2Al + 6HCl  2AlCl3 + 3H2 If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced? 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?

Question 1 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 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 NH3 # 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 NH3 N2 is the limiting reagent

Question 3 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 Al2O3 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 Al2O3 x 20.0 g O2 = 42.5 g Al2O3

Question 4 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)

Percent Yield Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100 x 100 96.7% =

Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100 x 100 17.8% =

Challenging question 2H2 + O2  2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100 x 100 92.9% =

More Percent Yield Questions The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2 What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

More Percent Yield Questions Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? Actual yield is given: 12.3 g O2 Next, calculate theoretical yield # g O2= 1 mol H2O 18.02 g H2O x 1 mol O2 2 mol H2O x 32 g O2 1 mol O2 x 14.0 g H2O 12.43 g = Finally, calculate % yield actual theoretical 12.3 g O2 12.43 g O2 = % yield = x 100% x 100% 98.9% =

Q2 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3  2KCI + 3O2 Actual yield is given: 107 g O2 Next, calculate theoretical yield # g O2= 1 mol KClO3 122.55 g KClO3 x 3 mol O2 2 mol KClO3 x 32 g O2 1 mol O2 x 300 g KClO3 117.5 g = Finally, calculate % yield actual theoretical 107 g O2 117.5 g O2 = % yield = x 100% x 100% 91.1% =

Q3 What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS Actual yield is given: 220 g FeS Next, calculate theoretical yield 1 mol FeS 1 mol Fe x 87.91 g FeS 1 mol FeS x # g FeS= 3.00 mol Fe 263.7 g = Finally, calculate % yield actual theoretical 220 g O2 263.7 g O2 = % yield = x 100 x 100 83.4% =

# g Fe2O3= 1 mol FeS2 119.97 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x 4FeS2 + 11O2  2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 1 mol FeS2 119.97 g FeS2 x 2 mol Fe2O3 4 mol FeS2 x 159.7 g Fe2O3 1 mol Fe2O3 x 300 g FeS2 199.7 g Fe2O3 = 1 mol O2 32 g O2 x 2 mol Fe2O3 11 mol O2 x 159.7 g Fe2O3 1 mol Fe2O3 x 200 g O2 181.48 g Fe2O3 = actual theoretical 143 g Fe2O3 181.48 g Fe2O3 = % yield = x 100% x 100 78.8% =

For more lessons, visit www.chalkbored.com 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 # g Cl2= 1 mol MnO2 86.94 g MnO2 x 1 mol Cl2 1 mol MnO2 x 70.9 g Cl2 1 mol Cl2 x 70 g MnO2 57.08 g Cl2 = 1 mol Cl2 4 mol HCl x 71 g Cl2 1 mol Cl2 x 3.5 mol HCl 62.13 g Cl2 = actual theoretical x g Cl2 57.08 g Cl2 = % yield = x 100% x 100 42% = 42 x 57.08 g Cl2 100 = x g Cl2 24 g Cl2 = For more lessons, visit www.chalkbored.com