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Chapter 12: Day 5
Ch12_stoic
22
STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
Mass reactant
StoichiometricfactorMoles
reactantMoles product
Mass product
Molar massgiven
Molar massUnknown
33
PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and O and HH22O are formed? What is the O are formed? What is the theoretical yield of products?theoretical yield of products?
PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and O and HH22O are formed? What is the O are formed? What is the theoretical yield of products?theoretical yield of products?
STEP 1STEP 1
Write the balanced Write the balanced chemical equationchemical equation
NHNH44NONO33 ---> --->
NN22O + 2 HO + 2 H22OO
44
454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) --> moles(454 g) --> moles
454 g • 1 mol
80.04 g = 5.68 mol NH4NO3
STEP 3 STEP 3 Convert moles reactant Convert moles reactant (5.68 mol) --> moles product(5.68 mol) --> moles product
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STEP 3 Convert moles reactant --> moles product
Relate moles NH4NO3 to moles product expected.
1 mol NH4NO3 --> 2 mol H2O
Express this relation as the STOICHIOMETRIC
FACTOR.
2 mol H2O produced1 mol NH4NO3 used
2 mol H2O produced1 mol NH4NO3 used
66
454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
= 11.4 mol H= 11.4 mol H22O producedO produced
5.68 mol NH4NO3 • 2 mol H2O produced1 mol NH4NO3 used
STEP 3 STEP 3 Convert moles reactant (5.68 Convert moles reactant (5.68 mol) --> moles productmol) --> moles product
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454 g of NH4NO3 --> N2O + 2 H2O 454 g of NH4NO3 --> N2O + 2 H2O
11.4 mol H2O • 18.02 g1 mol
= 204 g H2O
STEP 4 Convert moles product (11.4 mol) --> mass product
Called the THEORETICAL YIELD
ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY
PROBLEMS!!
ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY
PROBLEMS!!
88
0.21 mol AuCl3 = 64 g x 1molAuCl3
304 g AuCl3
Mole ratio = 3Cl2
2AuCl3
= 0.32 mol Cl2
X 71 g Cl2
1mol Cl2
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
Gases in Equations
The volume or amount of a gas in a chemical reaction can be calculated from
• the ideal gas law• mole-mole factors from the balanced
equation• molar mass
1010
IDEAL GAS LAWIDEAL GAS LAW
n is proportional to Vn is proportional to V(if T and P set)(if T and P set)
n is proportional to Pn is proportional to P(if V and T set)(if V and T set)
P V = n R TP V = n R T
1111
Mole ratio = 2SO3 = VOLUME ratio = 12L 2SO2
O2 = VOLUME ratio = 6L 2H2
SO2 = VOLUME ratio = 22.4L O2
4CO2 = VOLUME ratio = 14L 2C2H6
7O2 = VOLUME ratio = 3.5ft3 2C2H6
In = 2+7 Out = 4+6 out > in
Basic Chemistry Copyright © 2011 Pearson Education, Inc.12
1313
Gases and Gases and StoichiometryStoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.
1414
2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)
1.1 g of H2O2 is placed in a flask with a volume of 2.50 L.
What is the pressure of O2 at 25 oC? Of H2O?
Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V.
Strategy: Calculate moles of H2O2 and then moles of O2 and H2O. Finally, calc. P from n, R, T, and V.
15152 H2O2(liq) ---> 2 H2O(g) + O2(g)
SolutionSolution
1.1 g H2O2 • 1 mol34.0 g
0.032 mol1.1 g H2O2 • 1 mol34.0 g
0.032 mol
0.032 mol H2O2 • 1 mol O2
2 mol H2O2= 0.016 mol O20.032 mol H2O2 •
1 mol O22 mol H2O2
= 0.016 mol O2
1616Gases and Gases and StoichiometryStoichiometry
SolutionSolution
P of O2 = nRT/V
= (0.016 mol)(0.0821 L • atm/K •mol)(298 K)
2.50 L
P of O2 = nRT/V
= (0.016 mol)(0.0821 L • atm/K •mol)(298 K)
2.50 L
P of OP of O22 = 0.16 atm = 0.16 atm
1717
What is Pressue of H2O?
Could calculate as above. OR recall Avogadro’s hypothesis.
P n at same T and V
• 2HO2 = PRESSURE ratio X 0.16 atmO2
1O2
P of H2O = 0.32 atm
2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.
What volume, in L, of Cl2 gas at 1.20 atm and 27 °C is needed to completely react
with 1.50 g of aluminum?
2Al(s) + 3Cl2(g) 2AlCl3(s)
STEP 1 Calculate the moles of given using molar mass or ideal gas law.
1 mol of Al = 26.98 g of Al 1 mol Al and 26.98 g Al 26.98 g Al 1 mol Al
1.50 g Al x 1 mol Al = 0.0556 mol of Al 26.98 g Al
Basic Chemistry Copyright © 2011 Pearson Education, Inc.19
2Al(s) + 3Cl2(g) 2AlCl3(s)
STEP 2 Determine the moles of needed using a mole- RATIO
0.0556 mol Al x 3 mol Cl2 = 0.0834 mol of Cl2 2 mol Al
Basic Chemistry Copyright © 2011 Pearson Education, Inc.20
STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V.
T = 27 °C + 273 = 300. K
V = nRT = (0.0834 mol Cl2)(0.0821 L• atm/mol K)(300. K) P 1.20 atm
= 1.71 L of Cl2
Basic Chemistry Copyright © 2011 Pearson Education, Inc.21
What volume (L) of O2 at 24 °C and 0.950 atm is needed to react with 28.0 g of NH3?
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.22
STEP 1 Calculate the moles of given using molar mass or ideal gas law.
1 mol of NH3 = 17.03 g of NH3
1 mol NH3 and 17.03 g NH3
17.03 g NH3 Al 1 mol NH3
28.0 g NH3 x 1 mol NH3 = 1.64 mol of NH3
17.03 g NH3
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.23
STEP 2 Determine the moles of needed using a mole-mole factor.
5 mol of O2 = 4 mol of NH3
4 mol NH3 and 5 mol O2
5 mol O2 4 mol NH3
1.64 mol NH3 x 5 mol O2 = 2.05 mol of O2
4 mol NH3
Solution (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.24
STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law. To determine liters of gas, use the ideal gas law arranged to solve for V.
T = 24 °C + 273 = 297 K
Place the moles of O2 in the ideal gas law.
V = nRT =(2.05 mol)(0.0821 L• atm/mol K)(297 K)
P 0.950 atm
= 52.6 L of O2
Solution (continued)
Basic Chemistry Copyright © 2011 Pearson Education, Inc.25
What mass of Fe will react with 5.50 L of O2 at STP?
4Fe(s) + 3O2(g) 2Fe2O3(s)
1) 13.7 g of Fe
2) 18.3 g of Fe
3) 419 g of Fe
Learning Check
Basic Chemistry Copyright © 2011 Pearson Education, Inc.26
STEP 1 Calculate the moles of given using molar mass or ideal gas law. Use molar volume at STP to calculate moles of O2.
5.50 L O2 x 1 mol O2 = 0.246 mol of O2
22.4 L O2
STEP 2 Determine the moles of needed using a mole-mole factor. 4 mol of Fe = 3 mol of O2
4 mol Fe and 3 mol O2
3 mol O2 4 mol Fe 0.246 mol O2 x 4 mol Fe = 0.328 mol of Fe
3 mol O2
Solution
Basic Chemistry Copyright © 2011 Pearson Education, Inc.27
STEP 3 Convert the moles of needed to mass or volume using molar mass or ideal gas law.
1 mol of Fe = 55.85 g of Fe 1 mol Fe and 55.85 g Fe 55.85 g Fe 1 mol Fe 0.328 mol Fe x 55.85 g Fe = 18.3 g of Fe
1 mol Fe
Placing all three steps in one setup gives (STEP 1) (STEP 2) (STEP 3)
5.50 L O2 x 1 mol O2 x 4 mol Fe x 55.85 g Fe = 18.3 g of Fe 22.4 L O2 3 mol O2 1 mol Fe
Solution (continued)
