solving a stoichiometry problem
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Solving a Stoichiometry Problem. Balance the equation. Convert given to moles. Determine which reactant is limiting. Use moles of limiting reactant and mole ratios to find moles of desired product. Convert from moles to grams, molecules or Liters. Practice Test 2 (Ch. 9). - PowerPoint PPT PresentationTRANSCRIPT
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
1.1. Balance the equation.Balance the equation.
2.2. Convert given to moles.Convert given to moles.
3.3. Determine which reactant is limiting.Determine which reactant is limiting.
4.4. Use moles of limiting reactant and mole Use moles of limiting reactant and mole ratios to find moles of desired product.ratios to find moles of desired product.
5.5. Convert from moles to grams, molecules Convert from moles to grams, molecules or Liters.or Liters.
Practice Test 2 (Ch. 9)Practice Test 2 (Ch. 9)
1. How many moles of hydrochloric acid (HCl) would be produced from 4.18 moles of aluminum chloride (AlCl3)?
Use Mole Ratio (Balanced Equation)
4.18 mol AlCl3 = mol HClmol AlCl3
mol HClX
212.54
6
2 Al + 6 HCl 2 AlCl3 + 3 H2
4.18 mol
HClmol
Practice Test 2 (Ch. 9)Practice Test 2 (Ch. 9)
2. If you wanted to completely react 7.5 moles of hydrochloric acid (HCl), how many moles of Aluminum would be needed?
Use Mole Ratio (Balanced Equation)
7.5 mol HCl = mol Almol HCl
mol AlX
62.5
2
2 Al + 6 HCl 2 AlCl3 + 3 H2
7.5 mol Almol
3. How many grams of hydrochloric acid (HCl) would be needed to completely react 76.2 g of aluminum foil?
1
HHydrogen
1.01
17
ClChlorine
35.45
13
AlAluminum
26.98
Al + HCl AlCl3 + H2
g Al
mol HClmol Al
mol Al mol HCl
g HCl
Mole ratio (Balanced Eqn.)
X
1
6 2
Given: Want:
2
X X = g HCl
Convertto moles
(molar mass = 1 mole)
Convertto grams
(1 mole = molar mass)
1.01x 1
1.01
HCl
35.45x 1
+35.45= 36.46 g
26.98 2
6
1
Do the Calculation: 76.2 ÷ 26.98 x 6 ÷ 2 x 36.46 = 308.9234989 = 309
36.46309
3
76.2 g Al
4. How many grams of hydrochloric acid (HCl) would be needed to produce 15.0 dm3 of hydrogen gas?
1
HHydrogen
1.01
17
ClChlorine
35.45
Al + HCl AlCl3 + H2
dm3 H2
mol HClmol H2
mol H2 mol HCl
g HCl
Mole ratio (Balanced Eqn.)
X
1
6 2
Given: Want:
2
X X = g HCl
Convertto moles
(22.4 dm3 = 1 mole)
Convertto grams
(1 mole = molar mass)
1.01x 1
1.01
HCl
35.45x 1
+35.45= 36.46 g
22.4 3
6
1
Do the Calculation: 15.0 ÷ 22.4 x 6 ÷ 3 x 36.46 = 48.83035714 = 48.8
36.4648.8
3
15.0 dm3 H2
5. How many molecules of carbon disulfide (CS2) would be produced by completely reacting 8.00 grams of carbon (C) with excess sulfur dioxide gas (SO2)?
6
CCarbon12.01
C(s) + SO2(g) CS2(l) + CO(g)
g C
mol CS2mol C
mol C mol CS2
molecules CS2
Mole ratio (Balanced Eqn.)
X
1
2 1
Given: Want:
5
X X = molecules CS2
Convertto moles
(molar mass = 1 mole)
Convertto molecules
(1 mole = 6.02 x 1023)
12.01 5
1
1
Do the Calculation: 8.00 ÷ 12.01 ÷ 5 x 6.02 x 1023 = 8.01998 x 1022
6.02 x 1023
8.02 x 1022
4
8.00 g C
6. What volume (in Liters) of sulfur dioxide gas (SO2) is needed to produce 6.0 L of carbon monoxide gas (CO)?
C(s) + SO2(g) CS2(l) + CO(g)
L CO
mol SO2mol CO
mol CO mol SO2
L SO2
Mole ratio (Balanced Eqn.)
X
1
2
Given: Want:
5
X X = L SO2
Convertto moles
(22.4 L = 1 mole)
Convertto Liters
(1 mole = 22.4 L)
22.4 4
2
1
Do the Calculation: 6.0 ÷ 22.4 x 2 ÷ 4 x 22.4 = 3
22.43.0
4
6.0 L CO
7. What volume (in Liters) of ammonia gas (NH3) would be produced from 70.05 grams of nitrogen (N2) at STP
N2(g) + H2(g) NH3(g)
g N2
mol NH3mol N2
mol N2 mol NH3
L NH3
Mole ratio (Balanced Eqn.)
X
1
3
Given: Want:
1
X X = L NH3
Convertto moles
(molar mass = 1 mole)
Convertto Liters
(1 mole = 22.4 L)
28.02 1
2
1
Do the Calculation: 70.05 ÷ 28.02 x 2 x 22.4 = 112
22.4112.0
2
70.05 g N2
7
NNitrogen
14.01
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine:
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
a. The limiting reactantIdeal situation(From Bal.Eqn.)
Reactants
Compare the two reactants from balance equation tomoles given for each reactant in the problem.
C4H10
O2
=2.5 0.14
Moles given inProblem
C4H10
O2
=2
18=
0.15
1
0.14 is less than 0.15 so you don’t have enough butane which means
2.5 moles of butane, C4H10 is the limiting reagent.
13=
1
“I have” “I need”
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine:
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
b. The excess reactantIdeal situation(From Bal.Eqn.)
Reactants
Compare the two reactants from balance equation tomoles given for each reactant in the problem.
O2
C4H10
=18 7.2
Moles given inProblem
O2
C4H10
=13
2.5=
6.5
1
7.2 is more than 6.5 so you have “extra” moles of O2 making it the excess reagent.
2=
1
“I have” “I need”
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: c. The moles of carbon dioxide (CO2) produced.
Limiting
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine:
Start with the limiting reagent
2.5 mol C4H10 = ________ mol CO2x
c. The moles of carbon dioxide (CO2) produced.
Use Mole Ratio(Bal. Equation)
2
mol C4H10
mol CO2
8
10.
Identify what you want to find
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: d. How many grams of excess reactant left
over?
ExcessLimiting
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine:
Start with the limiting reagent
2.5 mol C4H10 = ________ mol O2x
d. How many grams of excess reactant left over ?
Use Mole Ratio(Bal. Equation)
13
mol C4H10
mol O2
2
16.25
Identify what you want to find1st find out how much excess gets used up.
Amount Used up
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: d. How many grams of excess reactant left over
?
ExcessLimiting
d. How many grams of excess reactant left over ?
ExcessReagent
(original amount)
mol O216.25
Amount Used up
18.0 mol O2
- = 1.75 mol O2
Excess molesLeft over
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)C
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: d. How many grams of excess reactant left
over?
ExcessLimiting
1.75 mol O2 = ________ g O2x
d. How many grams of excess reactant left over ?
Use molar mass(Periodic table)
1 mol O2
g O2 56
1 mole = Molar mass (Periodic table)
grams of excessreactant.
3256.0
3 sig.figs.
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: e. How many grams of water will theoretically
be produced fro this reaction?
Limiting
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine:
Start with the limiting reagent
2.5 mol C4H10 = ________ g H2Ox
e. How many grams of water will theoretically be produced fro this reaction?
Use Mole Ratio(Bal. Equation)
2
mol C4H10
mol H2O
10
225.25
Identify what you want to find
x1 mol H2O
18.02 g H2O
Theoretical Yield
1 mole = molar mass(Periodic Table)
Solving a Stoichiometry Solving a Stoichiometry ProblemProblem
8. If 2.5 moles of butane is mixed with 18.0 molesof O2, and the mixture is ignited, determine: f. Determine the % yield if the actual amount of
water produced during this reaction is 91.10 g.
ExcessLimiting
% Yield=Theoretical Yield
Actual Yield=
225.25 g
91.10 g
from previousProblem
X 100 X 100
= 40.4 %
2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (l)