Unit 6 ~ Stoichiometery (Chapter 9)

6-1 Mole – Mole Calculations (Section 9.1 – 9.2) To review, the coefficients of a balanced chemical equation can represent atoms, molecules, or moles. They DO NOT represent masses. N2 + 3 H2 → 2 NH3 1 molecule of N2 + 3 molecules of H2 → 2 molecules of ammonia 1 mole of N2 + 3 moles of H2 → 2 moles of ammonia

We can use the coefficients, then, to convert from moles of one substance to moles of another substance. For example,   N2 + 3H2 → 2NH3 1 mole 3 mole 2 mole 2 mole 6 mole 4 mole 0.5 mole 1.5 mole 1 mole

WOW THAT IS LIKE SO COOL!!!! N2 + 3H2 → 2NH3 1.35 moles of N2 will react completely with ? moles of H2? 1.35 mol N2 x 3 mole H2 = 4.05 mol of H2 1 mole N2 WOW THAT IS LIKE SO COOL!!!!  

How many moles of ammonia could be produced from 1.6 moles H2? Note how the conversion factor for a mole-mole calculation uses the coefficients! Practice: How many moles of ammonia could be produced from 1.6 moles H2? (When other chemicals aren’t mentioned in the question, you can assume that there is an “excess” of that chemical and it is not necessary to consider in the calculation.) N2 + 3H2 → 2NH3

N2 + 3H2 → 2NH3 1.6 mol H2 x 2 mole NH3 = 1.1 mol of NH3 3 mole H2 ? 1.6 mol H2 x 2 mole NH3 = 1.1 mol of NH3 3 mole H2 NOT SO BAD EH???????????????????????????

6-2 Mass – Mass Calculations (Section 9.3) One of the most common calculations in chemistry is to convert the mass of one substance into the mass of another substance. To do this, you first need a balanced chemical equation, like the combustion of propane:  C3H8 + 5 O2 → 3 CO2 + 4 H2O A problem might ask: How many grams of H2O are produced from the burning of 10.0 g C3H8?

? moles ? moles C3H8 + 5 O2 → 3 CO2 + 4 H2O 10g ? g The basic idea behind this type of “mass-mass” problem is to: 1) convert grams of C3H8 into moles of C3H8 using the molar mass 2) convert moles C3H8 into moles H2O using the coefficients 3) moles H2O into grams H2O using the molar mass

grams C3H8  moles C3H8  moles H2O  grams H2O Equation: C3H8 + 5 O2 → 3 CO2 + 4 H2O Grams: 10.0 g ? g Molar M: 44.06 g/mol 18.02 g As a dimensional analysis sequence: 10.0 g C3H8 x 1 mol C3H8 x 4 mol H2O x 18.02 g H2O 44.06 g C3H8 1 mol C3H8 1 mol H2O

Note: we could stop after step 1, 2, or 3 depending on what we are asked to calculate.   Practice: Diagram the entire problem for the combustion of 5.0 g CH4. Then calculate how much (g) carbon dioxide can be produced. Begin by writing out the complete balanced reaction keeping in mind your studies of the products of a combustion reaction.

CH4 + O2  CO2 + H2O 2 2 5g CH4 (1 mol CH4 / 16.05 g) = .31153 mol CH4 excess ?g 16.05 g/mol 44.01 g/mol 5g CH4 (1 mol CH4 / 16.05 g) = .31153 mol CH4 (1 mol CO2 /1 mol CH4) = .31153 mol CO2 (44.01g / 1 mol CO2) = Therefore 14 g with 2 sig figs. .31153 mol CH4 .31153 mol CO2 13.71g CO2