Ch. 9.1 & 9.2 Chemical Calculations

POINT > Define the mole ratio POINT > Use the mole ratio as a conversion factor POINT > Solve for unknown quantities using mole ratios and other conversion factors

A mole ratio is given by the coefficients of a balanced chemical equation 4 Al + 3 O 2  2 Al 2 O 3 From this equation you can derive 6 mole ratios: And the reciprocals of these.. 4 mol Al 3 mol O 2 4 mol Al 2 mol Al 2 O 3 3 mol O 2 2 mol Al 2 O 3

In stoichiometry you will use these ratios as conversion factors 4 Al + 3 O 2  2 Al 2 O 3 Ex. If you start with 2.5 mol Al, how much Al 2 O 3 could be produced? 2.5 mol Al x 2 mol Al 2 O 3 4 mol Al = 1.25 mol Al 2 O 3

In stoichiometry you will use these ratios as conversion factors 4 Al + 3 O 2  2 Al 2 O 3 Ex. If 12.7 mol of Al 2 O 3 are produced, how many mol oxygen are required? 12.7 mol Al 2 O 3 x 3 mol O 2 2 mol Al 2 O 3 = 19.1 mol O 2

N 2 (g) + 3H 2 (g)  2NH 3 (g) Determine the mole ratio of: Nitrogen to hydrogen Nitrogen to ammonia Hydrogen to ammonia 1 mol N 2 / 3 mol H 2 1 mol N 2 / 2 mol NH 3 3 mol H 2 / 2 mol NH 3 WB CHECK: Given the following equation

General formula for mole to mole problems: x mol G b mol W x a mol G = xb a mol W x = given value a, b = coefficients from equation G, W = Elements, Compounds

N 2 (g) + 3H 2 (g)  2NH 3 (g) How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 1.2 mol NH 3

Mass can’t be calculated directly from an equation of reactants and products, but we can convert mass to moles, then use mole ratio conversions

General formula for mass-mass problems: aGaG bWbW given quantity (g)wanted quantity (g)

N 2 (g) + 3H 2 (g)  2NH 3 (g) Ex. Determine the mass of ammonia produced by the reaction of 5.40g of hydrogen with an excess of nitrogen. 5.40g H 2 x 1 mol H g H 2 = 2.70 mol H mol H 2 x 2 mol NH 3 3 mol H 2 =1.80 mol NH 3

N 2 (g) + 3H 2 (g)  2NH 3 (g) Determine the mass of ammonia produced by the reaction of 5.40g of hydrogen with an excess of nitrogen mol NH 3 x 17.0 g NH 3 1 mol NH 3 = 30.6 g NH 3

N 2 (g) + 3H 2 (g)  2NH 3 (g) If 12.5 g of ammonia is produced, how many grams of nitrogen gas was required? 12.5 g NH 3 x 1 mol NH g NH 3 = mol NH mol NH 3 x 1 mol N 2 2 mol NH 3 =0.368 mol N mol N 2 x 28.0 g N 2 1 mol N 2 =10.3 g N 2

Other conversions work similar to mass-mass problems: Quantities must be converted to moles Mole ratio is used Moles converted to any other unit as required

General formula for solving aGaGbWbW given quantity wanted quantity

General formula for solving

2H 2 O (l)  2H 2 (g) + O 2 (g) How many molecules of oxygen are produced when 29.2g of water is decomposed by electrolysis? 4.88 x molecules O 2

You have many conversion factor relationships to utilize: Mole-Mole Mole-Massand Mass-Mole Particles-MoleandMole-Particle Volume-MoleandMole-Volume (coming soon)

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