1. Chemical Equations IV Stoichiometric Calculations

2. Coefficent in Balanced Equation A. Gives the relative number of molecules of reactants and products. Example: N 2(g) + 3H 2(g) --- > 2NH 3(g) One molecule of nitrogen gas molecules plus 3 molecules of hydrogen gas molecules forms 2 molecules of ammonia gas.

3. Coefficent in Balanced Equation B.Gives the relative number of moles of reactants and products. Example: N 2(g) + 3H 2(g) --- > 2NH 3(g) One mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas

4. Law of Conservation of Mass Mass of reactants = Mass of product Example: N 2(g) + 3H 2(g) --- > 2NH 3(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

5. 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

6. 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.

7. Example 1. Mole-Mole start 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: N 2(g) + O 2(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 N 2 x 2 mol NO 1 mol N 2 Step 3: Answer: 1.00 mol NO

8. 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.

9. Example 2. Mass-Mole start with mass – end with mole C 2 H 5 OH (ℓ) + 3O 2(g) ---- > 2CO 2(g) + 3H 2 O (ℓ) How many grams of CO 2 are produced when 3.00 g of C 2 H 5 OH is burned? 3.00 g C 2 H 5 OH X 1 mol C 2 H 5 OH X 2 mol CO 2 46.08 g C 2 H 5 OH 1 mol C 2 H 5 OH Answer: 0.130 mol CO 2

10. 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

11. Example 3. Mass – Mass start with mass – end with mass 2 C 8 H 18(ℓ) + 25 O 2(g) ---- > 16 CO 2(g) + 18 H 2 O (ℓ) How many grams of O 2 are needed to burn 2.00 g of C 8 H 18 ? 2.00 g C 8 H 18 X 1 mol C 8 H 18 X 25 mol O 2 x 32.0 g O 2 114.26 g C 8 H 18 2 mol C 8 H 18 1 mol O 2 Answer: 7.00 g O 2

12. 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

13. Example 4. Moles to Mass start with moles – end with mass 2 C 4 H 10(ℓ) + 13 O 2(g) ---- > 8 CO 2(g) + 10 H 2 O (ℓ) How many grams of CO 2 are produced when 1.72 x 10 -2 moles of C 4 H 10 burns in air? 1.72 x 10 -2 mol C 4 H 10 X 8 mol CO 2 X 44.01 g CO 2 2 mol C 4 H 10 1 mol CO 2 Answer: 3.03 g CO 2