Notes: StoichiometryStoichiometry is the study of amounts in chemical reactions. When doing stoichiometry problems, you must always begin with a balanced chemical reaction. You will then use the whole number coefficients as molar quantities for each compound, which will help to determine how much reactants are needed to produce a given amount of products.

Balanced Equations and Mole RatiosFrom a balanced equation, you can create mole ratios between two substances.

a) 2 H2 + 1 O2‡ 2 H2O b) 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2

Mole ratios: Mole ratios:2 moles H2 : 1 mole O2 1 mole Zn : 2 moles HCl2 moles H2 : 2 moles H2O 1 mole Zn : 1 mole H2

1 mole O2 : 2 moles H2 1 mole Zn : 1 mole ZnCl2

2 moles HCl : 1 mole H2

2 moles HCl : 1 mole ZnCl2

These mole ratios show the relationship between molar quantities of one compound to molar quantities of another compound

Stoichiometry Mole MapShows the relationship when converting between molar units AND between moles of one substance to moles of another substance.

use use22.4 L 22.4 L

use mole use

GFM ratio GFM

use use6.02 x 1023 6.02 x 1023

Liters of A Liters of B

Moles of A Moles of B

Atoms or Molecules

of A

Atoms or Molecules

of B

Grams of BGrams of A

1) 1 step stoichiometry calculations (mole – mole calculations)This is the simplest type of stoichiometry calculation. These are mole-mole calculations, in which you are trying to determine how many moles of one substance there are related to another substance. (Use a one-step dimensional analysis setup to solve.) Using the mole ratio of the two substances being asked about, you will create a setup as below to solve these problems.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2How many moles of H2 would be produced if there was 7.3 moles of HCl?

Given: 7.3 mole HCl x 1 mole H2 = 3.65 moles H2

1 2 moles HCl

mole ratio

Example 2:Given the following balanced equation: N2 + 3 H2‡ 2 NH3

How many moles of H2 are needed to produce 0.8 moles of NH3?

Given: 0.8 mole NH3 x 3 mole H2 = 1.2 moles H2

1 2 moles NH3

mole ratio

2) 2 – step stoichiometry calculationsIn a two-step calculation, you will be comparing moles of one substance to grams, liters or molecules of another substance in a balanced chemical equation. One step will be to convert the grams (or liters or atoms/molecule) to moles, and the other step will be to compare moles of one substance to moles of another substance using the mole ratios. This will require a two-step dimensional analysis setup.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2How many moles of H2 would be produced if there was 25.0 grams of Zn used?

Given: 25.0 g Zn x 1 mole Zn x 1 mole H2 = 0.382 moles H2

1 65.4 g Zn 1 mole Zn

GFM of Zn mole ratio

Example 2:Given the following balanced equation: N2 + 3 H2‡ 2 NH3

How many liters of NH3 would be produced if there was 5.5 moles of N2?

Given: 5.5 Moles N2 x 2 moles NH3 x 22.4 L NH3 = 246.4 L NH3

1 1 mole N2 1 mole NH3

mole ratio from mole map

Example 3:Given the following balanced equation: 2 KClO3‡ 2 KCl + 3 O2

How many moles of KClO3 would be needed to produce 2.60 x 1023 molecules of O2?

Given: 2.60 x 1023 molecules O2 x 1 mole O2 x 2 moles KClO3 = 0.29 moles KClO3

1 6.02 x 1023 molecules 3 mole O2

from mole map mole ratio

3) 3 – step stoichiometry calculationsIn a three-step calculation, you will be comparing grams/atoms/molecules/liters of one substance to grams, liters, atoms or molecules of another substance in a balanced chemical equation. You will follow the mole map road and use a three-step dimensional analysis setup.

Example 1:Given the following balanced equation: 1 Zn + 2 HCl ‡ 1 H2 + 1 ZnCl2

How many grams of ZnCl2 would be produced if there was 37.0 grams of Zn used?

Given: 37.0 grams Zn x 1 mole Zn x 1 mole ZnCl2 x 136.4 g ZnCl2 = 77.2 g ZnCl2

1 65.4 g Zn 1 mole Zn 1 mole ZnCl2

Example 2:Given the following balanced equation: 1 N2 + 3 H2‡ 2 NH3

How many molecules of N2 would be needed to produce 125.0 L of NH3?

Given: 125.0 L NH3 x 1 mole NH3 x 1 mole N2 x 6.02 x 1023 molecules = 1.68 x 1024 molecules

1 22.4 L NH3 2 mole NH3 1 mole N2

Example 3:Given the following balanced equation: 2 KClO3‡ 2 KCl + 3 O2

How many liters of O2 would be produced if the reaction began with 20.0 grams of KClO3?

Given: 20.0 g KClO3 x 1 mole KClO3 x 3 mole O2 x 22.4 L O2 = 5.5 L O2

1 122.6 g KClO3 2 mole KClO3 1 mole O2

Limiting Reactants and StoichiometryThe limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reactant, since the reaction cannot continue without it. If one or more other reactants are present in excess of the quantities required to react with the limiting reactant, they are described as excess reactants.The limiting reactant must be identified in order to calculate the percentage yield of a reaction, since the theoretical yield is defined as the amount of product obtained when the limiting reactant reacts completely. Given the balanced chemical equation, which describes the reaction, there are several equivalent ways to identify the limiting reactant and evaluate the excess quantities of other reactants.

Example #1: Limiting Reactant CalculationA 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

First, we need to create a balanced equation for the reaction: 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Next we can use stoichiometry to calculate how much product is produced by each reactant. NOTE: It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.

The reactant that produces the lesser amount of product in this case is the oxygen, which is thus the "limiting reactant." Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen).

We're not finished yet though. 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant remaining, subtract the amount that reacted from the amount in the original sample.

Example 2: Limiting Reactant Calculation90.0 g of FeCl3 reacts with 52.0 g of H2S. What is the limiting reactant? What is the mass of HCl produced? What mass of excess reactant remains after the reaction?

Percent Yield and StoichiometryThe theoretical yield is the maximum amount of product you would expect from a reaction based on the amount of limiting reagent. In practice, however, chemists don’t always obtain the maximum yield for many reasons. Sincechemists know that the actual yield might be less than the theoretical yield, we report the actual yield using percentyield, which tells us what percentage of the theoretical yield we obtained. This ratio can be very valuable to otherpeople who might try your reaction. The percent yield is determined using the following equation:

Percent Yield = Actual yield x 100Theoretical yield

Since percent yield is a percentage, you would normally expect to have a percent yield between 0% and 100%. Ifyour percent yield is greater than 100%, that probably means you calculated or measured something incorrectly.

Problem #1: What is the percent yield of the following reaction if 60.0 grams of CaCO3 is heated to give 15.0 grams of CaO?

CaCO3 --> CaO + CO2

Solution: Ideally, how many grams of CaO should be produced? First verify the equation is balanced; it is. Now determine the theoretical yield (in grams) of CaO, based on the amount of CaCO3 present.

60 grams CaCO3 x 1 mole CaCO3 x 1 mole CaO x 56.0 g CaO = 33.6 grams CaO1 100 grams CaCO3 1 mole CaCO3 1 mole CaO

So, ideally, 33.6 grams of CaO should have been produced in this reaction. This is the theoretical yield. However, the problem tells us that only 15 grams were produced. 15 grams is the actual yield. It is now a simple matter to find percent yield.

% Yield = 15.0 grams x 100 = 44.6 %33.6 grams

Problem #2: The following reaction is performed with 1.56 gram of BaCl2, and 1.82 grams of solid AgCl are produced.

BaCl2(aq) + 2 AgNO3 (aq) --> 2 AgCl(s) + Ba(NO3)2(aq)

What is the percent yield of the reaction?

1st: Determine the theoretical yield

1.56 grams BaCl2 x 1 mole BaCl2 x 2 mole AgCl x 143.3 g AgCl = 2.15 grams AgCl1 208.3 grams BaCl2 1 mole BaCl2 1 mole AgCl

So, ideally, 2.15 grams of AgCl should have been produced in this reaction. This is the theoretical yield. However, the problem tells us that only 1.82 grams were produced. 2.15grams is the actual yield. It is now a simple matter to find percent yield.

% Yield = 1.82 grams x 100 = 84.7 %2.15 grams