- The quantitative study of Chemical Change
- Most common Stoichiometry problem is the mass-mass problem or a "Given this amount of reactant, how much product will form?" problem
- Mass-Mass Problems:
- Steps:
- Write a balanced equation for the reaction
- Write the given mass on a factor-label form
- Convert mass of the reactant to moles of the reactant
- Convert moles of the reactant to moles of the product
- Convert moles of the product to grams of the product
- Solve
- Hint! Involves Molar Mass!
- Steps:
Practice Problem:
A.) If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as the coal burns. If a furnace burns an amount of coal containing 125 grams of FeS2, how much SO2 (an air pollutant) is produced?
1.) Write the balanced equation showing the formation of Iron (III) Oxide and sulfur Dioxide: d _FeS2 + _O2 -> _Fe2O3 + _SO2 = 4FeS2+ 11O2 -> 2Fe2O3 + 8SO2
2.) Write the mass information given in the problem:
125 grams FeS2 = ______________________________
3.) Convert grams of FeS2 to Moles of FeS2:
125 grams FeS2 = 1 mole FeS2 x ____________ c 1 12 grams FeS2
4.) Convert moles FeS2 (reactant) to moles of SO2 (product):
125 grams FeS2 = 1 mole FeS2 x 2 mole SO2 x ___________ z 1 120grams FeS2 1 mole FeS2
5.) Convert moles of SO2 to Grams of SO2:
125 grams FeS2 = 1 mole FeS2 x 2 mole SO2 x 64 grams SO2
1 120 grams FeS2 1 mole FeS2 1 mole SO2
6.) Solve: All units cancel each other out, except for the grams of SO2
125 = 1 x 2 x 64 SO2 = 133 grams SO2
1 120 1 1
A.) If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as the coal burns. If a furnace burns an amount of coal containing 125 grams of FeS2, how much SO2 (an air pollutant) is produced?
1.) Write the balanced equation showing the formation of Iron (III) Oxide and sulfur Dioxide: d _FeS2 + _O2 -> _Fe2O3 + _SO2 = 4FeS2+ 11O2 -> 2Fe2O3 + 8SO2
2.) Write the mass information given in the problem:
125 grams FeS2 = ______________________________
3.) Convert grams of FeS2 to Moles of FeS2:
125 grams FeS2 = 1 mole FeS2 x ____________ c 1 12 grams FeS2
4.) Convert moles FeS2 (reactant) to moles of SO2 (product):
125 grams FeS2 = 1 mole FeS2 x 2 mole SO2 x ___________ z 1 120grams FeS2 1 mole FeS2
5.) Convert moles of SO2 to Grams of SO2:
125 grams FeS2 = 1 mole FeS2 x 2 mole SO2 x 64 grams SO2
1 120 grams FeS2 1 mole FeS2 1 mole SO2
6.) Solve: All units cancel each other out, except for the grams of SO2
125 = 1 x 2 x 64 SO2 = 133 grams SO2
1 120 1 1
- Limiting Reactant
- The reactant that is completely consumed in the reaction
- Not present in sufficient quantities to react with all other reactants
- Reactant STOPS when the limiting reaction is completely consumed
- Remaining reactants are called "Excess Reactants"
- Amount of the product formed is determined by the "limiting reactant"
- Steps:
- Write balanced equation for the reaction
- Convert both reactant quantities to Moles
- Determine the moles of product that could be formed by each reactant
- The least amount in step 3 identifies the limiting reaction
- Use the number of Moles of the product to determine the mass product
- Write balanced equation for the reaction
- Steps:
- The reactant that is completely consumed in the reaction
Practice Limit reaction Problem:
B.) What mass of water can be proceed by 4 grams of Hydrogen Gas reacting with 16 grams of Oxygen Gas?
1.)Write the balanced equation:
2 H2 + O2 -> 2 H2O
2.)Cover both reactants to Moles:
a.) 4 grams H2 x 1 mole H2 =
1 2 grams H2
b.) 16 grams O2 x 1 mole O2 =
1 32 grams O2
3.) Use the mole ration from step 2 and determine the moles of water that could be formed by each reactant:
a.) 4 grams H2 x 1 mole H2 x 2 mole H2O = 2 mole H2O
1 2 grams H2 2 mole H2
b.) 16 grams O2 x 1 mole O2 x 2 mole H2O = 1 mole H2O
1 32 grams O2 1 mole O2
4.) Oxygen produces the least amount of water.
a.) 4 grams H2 x 1 mole H2 x 2 mole H2O = 2 mole H2O
1 2 grams H2 2 mole H2
b.) 16 grams O2 x 1 mole O2 x 2 mole H2O x 18 grams H2O = 18 grams H2O
1 32 grams O2 1 mole O2 1 mole H2O
B.) What mass of water can be proceed by 4 grams of Hydrogen Gas reacting with 16 grams of Oxygen Gas?
1.)Write the balanced equation:
2 H2 + O2 -> 2 H2O
2.)Cover both reactants to Moles:
a.) 4 grams H2 x 1 mole H2 =
1 2 grams H2
b.) 16 grams O2 x 1 mole O2 =
1 32 grams O2
3.) Use the mole ration from step 2 and determine the moles of water that could be formed by each reactant:
a.) 4 grams H2 x 1 mole H2 x 2 mole H2O = 2 mole H2O
1 2 grams H2 2 mole H2
b.) 16 grams O2 x 1 mole O2 x 2 mole H2O = 1 mole H2O
1 32 grams O2 1 mole O2
4.) Oxygen produces the least amount of water.
- 16 grams of oxygen can't produce as much water as 4 grams of Hydrogen
- 16 grams of oxygen will be used up in the reaction before 4 grams of Hydrogen (In other words)
- Oxygen is the "limiting" reactant
a.) 4 grams H2 x 1 mole H2 x 2 mole H2O = 2 mole H2O
1 2 grams H2 2 mole H2
b.) 16 grams O2 x 1 mole O2 x 2 mole H2O x 18 grams H2O = 18 grams H2O
1 32 grams O2 1 mole O2 1 mole H2O
- Percent Yield:
- The quantity of product that is calculated to form when all the limiting reactant is used up is called the Theoretical Yield
- The amount of product actually obtained in a reaction is called the Actual Yield
- The Actual yield is almost always less than the Theoretical Yield
- Formula: Percent Yield = Actual Yield X 100 c xx Theoretical Yield
- The quantity of product that is calculated to form when all the limiting reactant is used up is called the Theoretical Yield