stoichiometry ch. 9. the arithmetic of equations 9-1
TRANSCRIPT
Stoichiometry
Ch. 9
The Arithmetic of Equations
9-1
Stoichiometry
• Stoichiometry = the calculation of quantities in chemical reactions.
2H2 + O2 2H2O
Interpreting Chemical Equations• N2(g) + 3H2(g) 2NH3(g)
N2(g) 3H2(g) 2NH3(g)
Atoms
Molecules
Moles
(coefficient)
1 mol N2 3 mol H2 2 mol NH3
Molar Mass
(don’t use coefficient)
28g N2 2g H2 17g NH3
Practice!• H2(g) + I2(g) 2HI(g)
H2(g) I2(g) 2HI(g)
Atoms
Molecules
Moles
(coefficient)
Mass
(don’t use coefficient)
Practice!• H2(g) + I2(g) 2HI(g)
H2(g) I2(g) 2HI(g)
Atoms
Molecules
Moles
(coefficient)
1 mol H2 1 mol I2 2 mol HI
Mass
(don’t use coefficient)
2g H2 126.9g x 2 = 253.8g I2
126.9g + 1g =
127.9g HI
Chemical Calculations
9-2
Mole-Mole Calculations• N2(g) + 3H2(g) 2NH3(g)
1 mol N2 + 3 mol H2 2 mol NH3
• Mole Ratios = used to calculate between moles of reactants and products.– 6 possible mole ratio’s from equation above:
1 mol N2 2 mol NH3 3 mol H2
3 mol H2 1 mol N2 2 mol NH3
3 mol H2 1 mol N2 2 mol NH3
1 mol N2 2 mol NH3 3 mol H2
Mol A Mol B
• Calculation:
____mol A x mol B = _____mol B
mol A• Example:
How many moles of NH3 are produced when 0.60mol of N2 reacts with H2?
N2 + 3H2 2NH3
0.60mol N2 x 2 mol NH3 = 1.2 mol NH3
1 mol N2
Stoichiometric Calculations
• Atoms/Molecules
of A
• Mass of A
• Volume of A
• Atoms/Molecules
of B
• Mass of B
• Volume of B
x Mol B x Mol A
atoms A x 1 mol A
6.02x1023a A
Mass A x 1 mol A molar mass A
Volume A x 1 mol A 22.4L A
x 6.02x1023a B = atoms B 1 mol B
x molar mass B = mass B 1 mol B
x 22.4L B = volume B 1 mol B
Mole Ratio
Practice!
• How many molecules of O2 are produced from 29.2g of water?
2H2O(l) 2H2(g) + O2(g)
mass A mol A/B molec B29.2g H2O x 1 mol H2O x 1 mol O2 x 6.02x1023molec O2 =
18g H2O 2 mol H2O 1 mol O2
= 4.88x1023 molecules O2
More Practice!• How many liters of oxygen are needed to
produce 34.2g of SO3?
2SO2(g) + O2(g) 2SO3(g)
Mass (g) mol volume (liters)
34.2g SO3 x 1 mol SO3 x 1 mol O2 x 22.4L O2 80g SO3 2 mol SO3 1 mol O2
= 4.79 L O2
Percent Yield
9-3
Calculating the Percent Yield
• Theoretical Yield: amount of product a chemical reaction predicts by stochiometry.
• Actual Yield: is less than the theoretical yield due to lab techniques, equipment, etc.
• Percent Yield: ratio of the actual yield to the theoretical yield expressed as a percent.
– Shows efficiency of reaction
– Percent yield = actual yield x 100%
theoretical yield
Example
• A theoretical yield, using stoichiometry, of 4.55g of ammonia is produced. The actual yield from lab is 3.86g. What is the percent yield of this reaction?
– Percent yield = 3.86g x 100 = 84.8%
4.55g
Example• What is the theoretical yield of CaO if 24.8g CaCO3 is
used? What is the percent yield if 13.1g CaO is produced in lab?
CaCO3(s) CaO(s) + CO2(g)
24.8g CaCO3 ? g CaO massmolmass
1) Theoretical yield =
24.8g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56g CaO 100g CaCO3 1 mol CaCO3 1 molCaO
= 13.9g CaO
2) Percent yield = 13.1g CaO x 100% = 94.2%
13.9g CaO