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Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products
Stoichiometric Calculations 	From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)
Stoichiometric Calculations C6H12O6 + 6 O2  6 CO2 + 6 H2O 	Starting with 1.00 g of C6H12O6…  	we calculate the moles of C6H12O6… 	use the coefficients to find the moles of H2O… 	and then turn the moles of water to grams
Sample #1 CH4 +2O22H2O + CO2 How many moles of CH4 are needed to make 13 moles of water?
Sample #2 CH4 +2O22H2O + CO2 How many moles of water are made from 10g of Oxygen?
Sample #3 CH4 +2O22H2O + CO2 How many grams of water are produced from 100 moles of CH4?
Sample #4 CH4 +2O22H2O + CO2 How many grams of Oxygen are needed in order to produce 200 g of carbon dioxide?
Limiting Reactants 	The limiting reactant is the reactant present in the smallest stoichiometric amount
Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount In other words, it’s the reactant you’ll run out of first (in this case, the H2)
Limiting Reactants 	In the example below, the O2 would be the excess reagent
Theoretical Yield The theoretical yield is the amount of product that can be made In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures
Percent Yield Actual Yield Theoretical Yield Percent Yield =					x 100 	A comparison of the amount actually obtained to the amount it was possible to make

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Stoichiometric Calculations

  • 1. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products
  • 2. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)
  • 3. Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2O Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6… use the coefficients to find the moles of H2O… and then turn the moles of water to grams
  • 4. Sample #1 CH4 +2O22H2O + CO2 How many moles of CH4 are needed to make 13 moles of water?
  • 5. Sample #2 CH4 +2O22H2O + CO2 How many moles of water are made from 10g of Oxygen?
  • 6. Sample #3 CH4 +2O22H2O + CO2 How many grams of water are produced from 100 moles of CH4?
  • 7. Sample #4 CH4 +2O22H2O + CO2 How many grams of Oxygen are needed in order to produce 200 g of carbon dioxide?
  • 8. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount
  • 9. Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount In other words, it’s the reactant you’ll run out of first (in this case, the H2)
  • 10. Limiting Reactants In the example below, the O2 would be the excess reagent
  • 11. Theoretical Yield The theoretical yield is the amount of product that can be made In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures
  • 12. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100 A comparison of the amount actually obtained to the amount it was possible to make