ii. gas stoichiometry stoichiometry continued…. 1 mol of a gas=22.4 l at stp a. molar volume at...
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II. Gas StoichiometryII. Gas Stoichiometry
Stoichiometry Stoichiometry Continued…Continued…
Stoichiometry Stoichiometry Continued…Continued…
1 mol of a gas=22.4 Lat STP
A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
LITERSOF
SOLUTION
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
Molarity (mol/L)
B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem
How many grams of CaCO3 are req’d to
produce 9.00 L of CO2 at STP?
9.00 LCO2
1 molCO2
22.4 L CO2
= 40.2 g CaCO3
CaCO3 CaO + CO2
1 molCaCO3
1 molCO2
100.09g CaCO3
1 molCaCO3
? g 9.00 L
StoichiometryStoichiometryStoichiometryStoichiometry
III.AdjustinIII.Adjustingg
to Realityto Reality
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly
Limiting ReactantLimiting Reactant• bread
Excess ReactantsExcess Reactants• peanut butter and jelly
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of product
Excess ReactantExcess Reactant• added to ensure that the other
reactant is completely used up• cheaper & easier to recycle
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1g Zn
1 molZn
65.39g Zn
= 27.1 L H2
1 molH2
1 molZn
22.4 LH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
22.4L H2
1 molH2
0.90L
2.5 molHCl
1 L= 25 L
H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 27.1 L H2 HCl: 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
When 45.8 g of K2CO3 react with excess
HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
45.8 gK2CO3
1 molK2CO3
138.21 gK2CO3
= 49.4g KCl
2 molKCl
1 molK2CO3
74.55g KCl
1 molKCl
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
Theoretical Yield = 49.4 g KCl
% Yield =46.3 g
49.4 g 100 =93.7%
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g