.

1. TYPES
OF
STOICHIOMETRIC
CALCULATIONS
PINGOL, MA. KRISTELIZAH O
BEED 1-3 DAY

2. STOICHIOMETRY
Stoichiometry (sometimes called
reaction stoichiometry to
distinguish it form composition
stoichiometry) is the calculation of
quantitative (measurable)
relationship of the reactants and
product in a balanced chemical
reaction (chemicals).

3. “THERE ARE FOUR MAJOR
CATEGORIES OF STIOCHIOMETRY
PROBLEMS”

4. 1.Mole-Mole Problems
Problem: How many moles of HCl are needed to react with 0.87 moles of Al?
Step 1: Balance the Equation and calculate the ratios
Al + HCl  AlCl3 + H2
2Al + 6HCl  2AlCl3 + 3H2
2Al:6CHl (1:3) 2Al:2AlCl3 (1:1) 2Al:3H2 (1:1.5)
Step 2: Find the moles of the given
0.87 moles of aluminum are reacted with hydrochloric acid
 Step 3: Calculate the moles using the ratios
Moles HCL = 0.87molAl x 3molHCl/1molAl = 2.6mol HCl
0.87mol Al 3mol HCl =2.6 mol HCl
1mol Al

5. 2.Mass-Mass Problems
(Strategy: Mass g Mole g Mole g Mass)
Problem: How many grams of Al can be created decomposing 9.8g of Al2O3?
Step 1: Balance The Equation & Calculate the Ratios
Al2O3  Al + O2
2Al2O3  4Al + 302
2Al2O3 : 4Al (1:2) 2Al2O3 : 302 (1:1.5)
Step 2: Find the Mass of the Given
9.8g Al2O3 are decomposed
Step 3: Calculate the moles of the given (mol/g)
9.8g Al2O3 x (1mol Al2O3/102g Al2O3) = 0.096 mol Al2O3
9.8g Al2O3 1mol Al2O3
102g Al2O3
= 0. 096 mol Al2O3

6. Step 4: Calculate the moles using the ratios
0.096 mol Al2O3 x (2 mol Al/1 mol Al2O3) = 0.19 mol Al
0.096mol Al2O3 2mol Al
1mol Al2O3
= 0.19 mol Al
Step 5: Calculate the mass using the new moles
0.19 mol Al x (27 g Al / 1mol Al) = 5.1g Al
0.19mol Al 27g Al
1mol Al
= 5.1g Al

7. 3.Mass-Volume Problems
(Strategy: Mass g Mole g Mole g Volume)
Problem: How many liters of H2 are created from the reaction of 20.0g K?
Step 1: Balance The Equation & Calculate the Ratios
K + H2O  KOH + H2
2K + 2H2O  2KOH + H2
2K:2H2O (1:1) 2K:2KOH (1:1) 2K:1H2 (2:1)
Step 2: Find the Mass of the Given
20.0g K are used in the reaction
Step 3: Calculate the moles of the given (mol/g)
20.0g K x (1 mol K / 39g K) = 0.513 mol K
2O.Og K 1mol K
39g K
= 0.513mol K

8. Step 4: Calculate the mole using the ratios
0.51 mol K x (1mol H2 /2mol K) = 0.266mol H2
0.226 mol H2 22.4L H2
1mol H2
= 0.266mol H2
Step 5: Calculate the volume using the new moles
0.266mol H2 x (22.4L H2 /1mol H2) =5.75L H2
0.266mol H2 22.4L H2
1mol H2
= 5.75H2

9. 4.Volume-Volume Problems
Problem: How many liters of SO2 will be produced from 26.9L O2 ?
Step 1: Balance the equation and calculate the ratios.
S2 + O2  SO2
S2 + 2O2  2SO2
2O2:1S2 (2:1) 2O2:2SO2 (1:1)
Step 2: Find the volume of the given 26.9L O2

10. Step 3: Calculate the moles of the given
26.9L O2 x (1 mol O2 / 22.4L) = 1.20 mol O2
26.9L O2 1mol O2
22.4L
= 1.20 mol O2
Step 4: Calculate the moles using the ratios
1.20 mol O2 X (1mol SO2 / 1mol O2) = 1.20mol SO2
1.20 mol O2 1mol SO2
1mol O2
= 1.20mol SO2
Step 5: Calculate the volume using the new moles
1.20 mol O2 x (1mol SO2 / 1mol O2) X (22.4L / 1mol) =26.9L SO2
1.2O SO2 22.4L SO2
1mol SO2
= 26.9L SO2

11. THANK YOU
and
GOD BLESS! 