Stoichiometry

Defined

• Stoichiometry –The study of quantitative (measurable) relationships that exist in chemical formulas and chemical reactions.

• You must have:• 1. The correct formulas – “criss-cross” or

prefixes• 2. Balanced Equations – Five types,

verify law or conservation of mass

Analysis of a chemical reaction/equation

• Consider:

• N2H4 + 2H2O2 → N2 + 4H2O

• Same as saying…• 1mole N2H4 + 2moles H2O2 → 1mole N2 + 4 moles H2O

• The mole ratio is the “recipe” for the reaction.

Mole-Mole Problems

• Factor-Label Method – problem solving method based upon treating units in calculations as if they were algebraic factors.

• N2H4 + 2H2O2 → N2 + 4H2O

• Ex.1.4 moles of N2H4 gives how many moles of N2?

• 1.4 moles N2H4 x mole N2 = moles N2

• mole N2H4

Another Example

• 3Zn + 2H3PO4 → Zn3(PO4)2 + 3H2

• How many moles of Zn3(PO4)2 will be produced from 2.18 moles of H3PO4?

• 2.18 moles H3PO4 x moles Zn3(PO4)2

• mole H3PO4

• 1.09 moles Zn3(PO4)2

Verifying the Law of Conservation of Matter

• 2H2 + O2 → 2H2O

• Calculate the mass of the reactants and the mass of the products

• 2 mol H2 x g H2 + 1 mol O2 x g O2 = 36.04g

• 1 mol H2 1 mol O2

• Or 2 mol H2O x g H2O = 36.04g

• 1 mol H2O

Mass – Mass Problems

• Steps • Find molar mass of both substances• change mass to moles of given• use molar bridge (moles of given moles of

unknown)• change moles to mass of unknown• ***Remember, coefficients in the balanced

equation represent the relative number of moles of reactants and products!***

Mass-Mass Problems

• What mass of water is produced from 1.5 grams of glucose?

• C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

• 1st. change mass to moles using molar mass of glucose. (180 g/mol)

• 2nd. Use molar bridge to change from moles of given to moles of unknown.

• 3rd change to grams of water

The example of Mass-Mass

• C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

• 1.5g C6H12O6 to g of water.

• 0.9g H2O

Another example

• Ex. What mass of aluminum oxide is produced when 2.3 g of aluminum reacts with iron (III) oxide?

• 4.3g Al2O3

Mass-Volume problems

• 1st mass of given

• 2nd moles of given (molar mass)

• 3rd moles of unknown (molar ratio)

• 4th volume of gaseous unknown

• If I have 125 g of Al2O3 how many L of Al do I have @ STP?

Another Example

• Find the mass of aluminum required to produce 1.32L of H2 gas @ STP

• 2Al + 3 H2SO4 → Al2(SO4)3 + 3 H2

•

Volume-volume problems

• Same as mole-mole just using volumes instead!

• If I have 15.5 L of N2 gas, how many L of H2 will react?

• N2 + 3 H2 → 2 NH3

11-3 Limiting Reactants and Percent Yield

• When chemicals combine, they are usually in nonstoichiometric proportions. That means there will be a limiting reactant.

• Limiting Reactant will be completely used up in the reaction.

• The other, leftover amount is said to be in excess.

• **The quantities of products formed in a reaction are always determined by the quantity of limiting reactant. **

LR continued

• To find the limiting Reactant:

• 1st solve 2 separate mass-mass problems.

• 2nd see which number is smaller; this is your limiting reactant.

Example of LR

• Ex. 3.5 g of Cu is added to 6.0 g silver nitrate. Find the limiting reactant.

• (Note: you can calculate the mass of either product, use the easier one to find the molar mass!)

• Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag

• You can also determine how much of the substance in excess is left over.

• 1st determine the LR and then use the amount of the LR and use that to find the mass of the other substance.

• Ex. 2.0 g of NaHCO3 reacts with 0.5 g of H3C6H5O7 (citric acid) which is the LR and how much of the other will be left over? What volume of CO2 will be produced?

example

• 3 NaHCO3 + H3C6H5O7 → 3 CO2 + 3 H2O + Na3C6H5O7

Percent yield

• Percent yield = actual yield x 100

• Expected

• Actual = what you got in lab

• Expected = what it was supposed to be

Example

• A piece of copper with a mass of 5.00g is placed in a solution of AgNO3 . The silver metal produced has a mass of 15.2g. What is the percent yield for this rxn?

• 1st calculate the mass of the expected silver.

• Cu + 2AgNO3 2 Ag + Cu(NO3)2

Example continued