Chemistry Unit 6

Stoichiometry (USE SIG FIGS)

branch of chemistry that deals with the quantities of substances that enter into and are produced by chemical reactions


-Balance the equation

-Get to moles

-Mole to mole ratio

ex.)2Fe + 3CO2 = 1Fe2O3 + 3CO (the equation is balanced). 2 moles of Fe = 3 moles CO2 or 1 mole Fe2O3 = 3 mols of CO

-Get to desired unit

Example)How many water molecules will be produced if 56.8 grams of phosphoric acid is reacted with calcium hydroxide?

3Ca(OH)2 + 2H3PO4 = 1Ca3(PO4)2 + 6H2O 56.8gH3PO4 x (1moleH3PO4/98gH3PO4) x (6molH2O/2molH3PO4) x (6.02 x 1023/1molH2O) = 1.04 x 1024

*For Stoichiometry, you will need to: write equations and predict products, balance equation, use the factor label method, convert units (molar mas, STP, Avagadro’s #), use sig figs


1.Theoretical Yield

2.Actual Yield/experimental Yield

3.Percent Yield

1.maximum amount of product that CAN be produced from given amounts of reaction. (thus the theoretical yield is almost always greater than the actual yield. 2.measured amount of product obtained from a reaction (given in problem)

3.ratio of actual yield to theoretical yield (in the same unit) times 100 (the closer you are to 100 in the % yield, the more accurate you are)

Example)C6H6 + Cl2 = C6H5Cl + HCl…When 36.8g of benzene reacts with Cl2, the actual yield of chlorobenzene(C6H5Cl) is 38.8g.

36.8gC6H6 x (1moleC6H6/78gC6H6) x (1molC6H5Cl/1moleC6H6) x (112.5gC6H5Cl/1molC6H5Cl) = 53.1gC6H5Cl(this is the theoretical yield)

Percent yield: (38.8g(actual)/53.1g(theoretical)) x 100 = 73.1%C6H5Cl

Info Related to Stoichiometry

The coefficients of a balanced equation tell us the relative number of moles of reactants and products. All stoichiometric calculations begin with a balanced equation. Balanced equations are necessary to attain a mole to mole ratio. In mass calculations, the molar mass is needed to convert mass to moles. In volume calculations 22.4L is needed to convert liters to moles. In atoms or molecule calculations, Avogadro’s # is used to convert atoms to moles

1.Limiting Reagents/Reactants

2.Excess Reagents/Reactants

1.the reactant in the equation that limits how much product can be made(reaction stops when limiting reagent is used up)

2.the reactant in the equation that is not completely used up in the reaction(reactant is not used up because it has nothing left to react with) Ex.)If you had t car bodies and 48 tires, you could only make 8 cars with 16 tires left over in excess. No matter how many tires you have, only 8 cars can be made because there are only 8 car bodies. Car bodies are the “limiting reagent” while tires are the “excess reagent”

Steps for Determining the Limiting and Excess Reactants

Always start with a balanced equation. Use stoichiometry to calculate how much product is produced by each reactant. (It doesn’t matter which product is chosen but the SAME product MUST be used for BOTH reactants so that the amounts can be compared). The reactant that produces the lesser amount of product is the limiting reagent. To find the excess reactant, calculate how much of the excess reagent actually did react with the limiting reagent and subtract from the amount in the original sample.


Example)If 4.95g of ethylene(C2H4) are combusted with 3.25g of oxygen, what is the limiting reagent? How many grams of each product are formed? What mass of the excess reactant remains once the reaction is complete?

C2H4 + 3O2 = 2CO2 + 2H2O – 4.95gC2H4 x (1molC2H4/28gC2H4) x (2molCO2/1moleC2H4)x (44gCO2/1molCO2) = 15.6gCO2…C2H4 is the excess reagent

– 3.25gO2 x (1molO2/32gO2) x (2molCO2/3molO2) x (44gCO2/1molCO2) = 2.98gCO2…O2 is the limiting reagent

– 3.25gO2 x (1molO2/32gO2) x (2molH2O/3molO2) x (18gH2O/1molH2O) = 1.22gH2O is produced

– 3.25gO2 x (1molO2/32gO2) x (1molC2H4/3molO2) x (28gC2H4/1molC2H4) = .948gC2H4…4.95

– .948 = 4.00gC2H4 of excess reagent is left over after the reaction is complete

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