Chemical Equations Chemical Equations IVIV

Chemical Equations Chemical Equations IVIV

Stoichiometric CalculationsStoichiometric Calculations

Coefficent in Balanced Equation

A. Gives the relative number of molecules of reactants and products.

Example: N2(g) + 3H2(g) --- > 2NH3(g)

One molecule of nitrogen gas molecules plus 3 molecules of hydrogen gas molecules forms 2 molecules of ammonia gas.

Coefficent in Balanced Equation

B. Gives the relative number of moles of reactants and products.

Example: N2(g) + 3H2(g) --- > 2NH3(g)

One mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas

Law of Conservation of Mass Mass of reactants = Mass of product

Example: N2(g) + 3H2(g) --- > 2NH3(g)

2(14.0 g/mol) + 6(1.01 g/mol) = 2(14.0g/mol) + 6(1.01 g/mol)

34.06 g/mol = 34.06 g/mol

Steps to a Correct Answer

1. Write a balanced equation.2. Write a dimensional analysis set-up with labeled values. Start with given

value in problem.3. Don’t do work in parts. I want to see

one step.4. Calculate answer

Stoichiometry I

Type:

Mole-Mole

moles of ---> moles of reactant product

• Use coefficients in balanced equation

• Put moles of chemical you’re solving for in the numerator.

Example 1. Mole-Molestart with mole – end with mole

How many moles of nitric oxide (nitrogen monoxide) are produced when 0.500 moles of nitrogen reacts with excess nitrogen?

Step 1: N2(g) + O2(g) --- > 2NO(g)

Step 2: Start with the given value. Write mole ratio using coefficients in balanced equation. Put the number of moles of the substance you’re solving for in the numerator.

0.500 mol N2 x 2 mol NO 1 mol N2

Step 3: Answer: 1.00 mol NO

Stoichiometry II

Type: Mass – Mole

mass of ---> moles of reactant product

• Change mass of chemical given to moles using it’s molar mass.

• Mole ratio from balanced equation. Put moles of chemical you’re solving for in numerator.

Example 2. Mass-Molestart with mass – end with mole

C2H5OH(ℓ) + 3O2(g) ---- > 2CO2(g) + 3H2O(ℓ)

How many grams of CO2 are produced when 3.00 g of C2H5OH is burned?

3.00 g C2H5OH X 1 mol C2H5OH X 2 mol CO2

46.08 g C2H5OH 1 mol C2H5OH

Answer: 0.130 mol CO2

Stoichiometry III

Type:

Mass – Mass

mass of ---> mass of reactant product

• Change mass to moles using molar mass

• Use mole ratio from balanced equation. Put moles of chemical you’re solving for in the numerator.

• Change moles to mass using molar mass

Example 3. Mass – Massstart with mass – end with mass

2 C8H18(ℓ) + 25 O2(g) ---- > 16 CO2(g) + 18 H2O(ℓ)

How many grams of O2 are needed to burn 2.00 g of C8H18?

2.00 g C8H18 X 1 mol C8H18 X 25 mol O2 x 32.0 g O2

114.26 g C8H18 2 mol C8H18 1 mol O2

Answer: 7.00 g O2

Stoichiometry IV

Type:

Moles – Mass

moles of ---> mass of reactant product

• Change mass of given chemical to moles using molar mass

• Use mole ratio from balanced equation putting moles of chemical you’re solving for in numerator.

• Change moles to mass using molar mass

Example 4. Moles to Massstart with moles – end with

mass

2 C4H10(ℓ) + 13 O2(g) ---- > 8 CO2(g) + 10 H2O(ℓ)

How many grams of CO2 are produced when 1.72 x 10-2 moles of

C4H10 burns in air?

1.72 x 10-2 mol C4H10 X 8 mol CO2 X 44.01 g CO2

2 mol C4H10 1 mol CO2

Answer: 3.03 g CO2