Chapter Chapter 1212

StoichiometStoichiometryry

StoichiometryStoichiometry—the calculation of

quantities in chemical reactionsCalculating how much of each reactant

is needed and how much of each product is formed.

Like a recipe: you need a certain amount of ingredients to get a certain amount of product.

Chemical Equations: What can you tell?

1N2 + 3H2 2NH3The coefficients in a Balanced Chemical

Equation tell the # of particles of each reactant & product

1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3.

OR The coefficients can represent the # of moles of each reactant & product.

1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

3 Steps of Stoichiometry1. Convert to moles (from particles, mass, or

volume).

2. Do a mole to mole conversion using the balanced chemical equation

3. Convert moles to desired unit (particles, mass, or volume).

1 Mole 1 MoleBCE

1. 2. 3.

Stoichiometry Example

1N2 + 3H2 2NH3

If 2.61 L of H2 gas reacts with the nitrogen in the air, how many grams of NH3 will be produced?

2.61 L H2 x 1 mol H2_22.4 L H2

2 mol NH3 3 mol H2

17.0 g NH3_ 1 mol NH3

= 1.32 g NH3

xx

Given 1 Mole BCE 1 Mole

Percent Yield In a lab, the amount of product formed is often less than expected based on the balanced chemical equation.

Theoretical Yield—the amount of product that should be formed (from stoichiometry problem)

Actual Yield—the amount of product actually formed in a lab (found in an experiment).

Percent Yield—the ratio of actual to theoretical yield. Percent Yield = actual yield x 100

theoretical yield

Example 24.8g calcium carbonate is decomposed by heating.

13.1g CaO is actually produced. What is the percent yield?

CaCO3 CaO + CO2

Actual Yield: 13.1g CaO Theoretical Yield: (use stoichiometry)

24.8g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56.1 g CaO

100.1 g CaCO3 1 mol CaCO3 1 mol CaO

=13.9 g CaO % Yield = 13.1 x100

13.9

Limiting Reagents

Limiting Reagent—limits or determines the amount of product that can be formed in a reaction.

Excess Reagent—the reactant that is not completely used up in a reaction.

Limiting Reactants video

5N2 and 9H2 6NH3 and 2N2

Hydrogen is Oxygen is limiting excess

Hydrogen islimiting

NitrogenNitrogen is excess

Solving for a Limiting Reagent:

1.1. Use stoichiometry to convert each Use stoichiometry to convert each reactant (individually) into product.reactant (individually) into product.

2.2. The reactant that gives you the least The reactant that gives you the least product is the product is the limitinglimiting reagent. reagent.

3.3. The reactant that gives the most The reactant that gives the most product is the product is the excess excess reagent.reagent.

Example: Nickel replaces silver from silver nitrate in solution according to the

following equation: 2AgNO3 + Ni → 2Ag + Ni(NO3)2 If you have 22.9 g of Ni and 112 g of AgNO3 ,what mass of nickel(II) nitrate would be produced?

22.9g Ni x 1 mol Ni x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 = 71.3 g Ni(NO3)2

58.7g Ni 1 mol Ni 1 mol Ni(NO3)2

112 g AgNO3x1 mol AgNO3 x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 =60.2g Ni(NO3)2

169.9g AgNO3 2 mol AgNO3 1 mol Ni(NO3)2

AgNO3 is limiting & 60.2 g Ni(NO3)2 can be produced