Unit 9 part 2: Stoichiometry. 9.2.1 Stoichiometry What is Stoichiometry? The study of the quantitative relationships that exist in chemical formulas and.

Unit 9 part 2:

Stoichiometry

9.2.1 Stoichiometry

What is Stoichiometry?The study of the quantitative

relationships that exist in chemical formulas and reactions.

• Stoicheion- element• Metron- measure

Interpreting Balanced Chemical Equations

To balance chemical equations we use coefficients- whole numbers in front of the formula.

These coefficient represent MOLE ratios for the substances in the rxn.

Ex.- N2H4 +2H2O2 → N2 + 4H2O

means…

1mol N2H4 +2mol H2O2 → 1mol N2 + 4mol H2O

Mole-Mole Problems

Knowing that the coefficients show MOLE RATIOS… we can now solve problems relating moles of one substance to moles of another.

ExN2H4 +2H2O2 → N2 + 4H2O

How many mols of N2H4 are needed to react with 23 mols of H2O2 ?

More mol- mol practice

  2S  +  3O2 →  2SO3

How many moles of S are needed to react with 2.5 mols of O2?

Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O

How many moles of Cu(NO3)2 are produced from 12.3 moles of HNO3?

Verifying the Law of Conservation of Matter

Now we can PROVE balanced equations obey the law of conservation of matter.

2H2 + O2 → 2H2O

We do this by calculating the mass of the reactant and comparing them to the mass of the products.

H- 2molX 2g = 4g plus O- 1mol X 32g= 32g equals 36g

H20- 2mol X 18g = 36g

9.2.2: Solving Stoichiometry Problems

There are 3 types of Stoichiometry problems:1. Mass-Mass problems2. Mass-Volume problems3. Volume-Volume problems

In general, every problem will be solved in 3 steps:Quantity of given →mols of given →mols of unknown → quantity of unknown

Mass-Mass Problems

In these problems, you will be given the mass of one substance and asked to solve for the mass of another substance.

Your mantra is: Mass to moles, moles to moles, moles to

mass. You will need a balanced chem. equation,

the molar mass of the known and the unknown substance to solve.

Mass-mass practice

2H2 + O2 → 2H2OHow many grams of water will be produced

from 23.5 g of oxygen?Mass to moles, moles to moles, moles to mass

Convert g to moles of O2, then mol O2 to mol H2O, and finally, mols of H2O to g of H2O.

More practice Zn + H2SO4 → ZnSO4 + H2

How many g of H2SO4 will react with 9.5 g of Zn?

HCl + NaOH → NaCl + H2OHow many g of NaCl will be produced from 51.2 g of

NaOH?

2Mg + O2 → 2MgOHow many g of Mg are needed to produce 12.3g

MgO?

Mass-Volume ProblemsIn these problems you will be given the MASS of

one substance and asked to find the VOLUME of a gas. You will be going from grams to liters (g → L).

Your Mantra is: Mass to moles, moles to moles, moles to

volume.

To convert from mols to volume you will use the molar volume (22.4L/ mol).

Mass-Volume practice

2H2 + O2 → 2H2OHow many liters of hydrogen will be required to

produce 23.5 g of water?Mass to moles, moles to moles, moles to

volume. Convert g to moles of H2O, then mol H2O to

mol H2, and finally, mols of H2 to L of H2.Remember, use the molar volume: 22.4 L/mol

More practice Zn + H2SO4 → ZnSO4 + H2

How many L of H2 will be produced from 9.5 g of Zn?

Sn + 2HF → SnF2 + H2

How many g of HF are needed to react with 30.0L of H2?

NH4NO3 → N2O + 2H2O How many L of N2O will be produced from 52.6 g NH4NO3?

2Mg + O2 → 2MgO How many g of MgO are needed to produce 12.3L O2?

Volume-Volume Problems

In these problems you will be given the VOLUME of one gas and asked to find the VOLUME of a different gas.

Your Mantra is: VOLUME to moles, moles to moles, moles to

volume.

To convert from mols to volume you will use the molar volume (22.4L/ mol).

Volume - Volume practice

2H2 + O2 → 2H2OHow many liters of hydrogen will be required to

react with 23.5 L of O2?volume to moles, moles to moles, moles to

volume. Convert L to moles of O2, then mol O2 to mol

H2, and finally, mols of H2 to L of H2. Remember, use the molar volume: 22.4 L/mol

More practice

Sn + 2HF → SnF2 + H2

How many L of HF will be required to form 3.00L of H2?

N2 + 3H2 → 2NH3

How many L of N2 will be required to react with 5.2 L H2?

9.3.3 Limiting Reactants and Percent Yields

The amount of product made depends on how much reactants are available. Sometimes, one of the reactants limits the number of products made… How many bikes can be made from 11 bike frames but only

7 tires? Did the # of bike frames or the # of tires limit how many

bikes were made? The reactant that limits the amount of product

formed is called the limiting reactant. The reactant that is not completely used up is

called the excess reactant.

Identifying Limiting Reactants

There are 3 general steps to identifying the limiting reactant.1. Begin with a balanced chemical equation.2. Calculate the amount of product formed by

EACH of the reactants. (pick only one product)3. The reactant that produces the least amount

of product is the limiting reactant.

Limiting Reactant Problems

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

Which is the limiting reactant if you have 6.0g Cu and 12.5g AgNO3 ? (hint: solve for Ag) You will perform 2 mass-mass problems.

• 1. mass of Cu to mass of Ag

• 2. mass of AgNO3 to mass of Ag.

Percent Yield

The amount we calculate and what we actually make as the product are not necessarily the same amount.

Expected yield – amount of product that should be produced based on calculation.

Actual yield – the amount ACTUALLY obtained from the reaction (in lab).

We need to know how much of the expected was made during the reaction. …did we only make 5% or 55%?

Percent Yield – what % of the predicted amount was actually made.

% yield = (actual yield ÷ expected yield) X 100%

Percent Yield practice problems

You burn Mg in O2 and produce 6.5g MgO. You expected to make 8.2g of MgO. What is your percent yield?

Determine the %yield for 3.74g Na and excess O2 if 5.34g of Na2O2 is recovered?

We don’t know the expected yield so you will have to calculate it first, then solve %yield.