# Reaction Stoichiometry.

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Reaction Stoichiometry

Introduction Reaction stoichiometry involves the mass relationships
between reactants and products in a chemical reaction. It is based on chemical equations similar to the ones studied in the last section. All reaction stoichiometry calculations start with a balanced equation. You will need to be familiar with gram/mole relationships as studied earlier this year.

The mole enables chemists to move from the microscopic world of atoms and molecules to the real world of grams . Stoichiometry problems are classified between the information given in the problem and the information you are expected to find, the unknown. The given and the unknown may be expressed in grams or moles. The masses in the reaction are usually expressed in grams. Definition of mole: mole of a substance = grams of substance/MW of substance

You will need to use: molar ratios in a balanced equation. molar masses of reactants and products. balancing equations. conversions between grams and moles.

Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound. All stoichiometry problems use mole ratios.

Example Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) 2 Mg(s) + O2(g) MgO(s) (balanced) Mole Ratios: 2 : 1 : 2

Stoichiometry (working with ratios)
Ratios are found within a chemical equation. 2HCl + Ba(OH)2  2H2O BaCl2 1 1 coefficients give MOLAR RATIOS 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

Practice Problems 1) N2 + 3 H2 ---> 2 NH3
Write the mole ratios for N2 to H2 and NH3 to H2.

Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = grams per mole H2O = grams per mole Ca(OH)2 = grams per mole

Review: Chemical Equations
C2H5OH + 3O2 ® 2CO2 + 3H2O reactants products 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Types of Stoichiometry Problems

Problem Type 1: Given and unknown quantities are in moles
Amount in moles of known substance Amount in moles of unknown substance How many moles of lithium hydroxide are required to react with 20 moles of CO2

20 mol CO2 x 2 mol LiOH / 1 mol CO2 = 40 mol LiOH
Given: amount of CO2 = 20 moles Unknown: amount of LiOH in moles Amount of CO2 in moles Amount of LiOH in moles mol CO2 x mol LiOH / mol CO2 = mol LiOH 20 mol CO2 x 2 mol LiOH / 1 mol CO2 = mol LiOH mole ratio

Mole – Mole Conversions
Problem Type 1 Mole – Mole Conversions When N2O5 is heated, it decomposes: 2N2O5(g)  4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g)  4NO2(g) + O2(g) 4.3 mol ? mol Units match

Mole – Mole Conversions
When N2O5 is heated, it decomposes: 2N2O5(g)  4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g)  4NO2(g) + O2(g) 4.3 mol ? mol Units match 4.3 mol N2O5 = moles NO2 8.6

2N2O5(g)  4NO2(g) + O2(g) 4.3 mol N2O5 = mole O2 2.2
b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g)  4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = mole O2 2.2

Problem Type 1: Given and unknown quantities are in moles
Amount in moles of known substance Amount in moles of unknown substance How many moles of ammonia are produced when 6 moles of hydrogen gas react with an excess of nitrogen gas.

Given: amount of H2 = 6 moles
Unknown: amount of NH3 in moles Amount of H2 in moles Amount of NH3 in moles (mol H2) x (mol NH3 / mol H2) = mol NH3 (6 mol H2) x (2 mol NH3 / 3 mol H2) = 4 mol NH3 Mole ratio

Problem Type 2: Given amount is in moles and unknown quantity is in grams
Amount in moles of known substance Amount in grams of unknown substance Problem Type 3: Given amount is in grams and unknown quantities are in moles Amount in grams of known substance Amount in moles of unknown substance

In plants when carbon dioxide reacts with water it
Problem Type 2 mole ↔ gram In plants when carbon dioxide reacts with water it produces glucose and oxygen: 6CO2 + 6H2O(l)  C6H12O6(s) + 6O2(g) How many grams of C6H12O6 is produced when 3.0 mol of water react with carbon dioxide? 6CO2 + 6H2O  C6H12O O2 3.0 mol ? grams Units match 3.0 mol H2O = g C6H12O6 90

In plants when carbon dioxide reacts with water it
Problem Type 2 mole ↔ gram In plants when carbon dioxide reacts with water it produces glucose and oxygen: 6CO2 + 6H2O(l)  C6H12O6(s) + 6O2(g) How many grams of CO2 is needed to react with 3.0 mol of water? 6CO H2O  C6H12O O2 ? grams 3.0 mol Units match 3.0 mol H2O = g CO2 132

When magnesium burns in air, it combines with oxygen
Problem Type 2 mole ↔ gram When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation: 2Mg + O2(g)  2MgO(s) How many grams of MgO is produced from 2.0 mol of magnesium? 2Mg O2  2MgO 2.0 mol ? grams Units match 2.0 mol Mg = g MgO 80

When N2O5 is heated, it decomposes: 2N2O5(g)  4NO2(g) + O2(g)
Problem Type 3 gram ↔ mole When N2O5 is heated, it decomposes: 2N2O5(g)  4NO2(g) + O2(g) a. How many moles of N2O5 were used if 210g of NO2 were produced? 2N2O5(g)  4NO2(g) + O2(g) ? moles 210g Units match 210 g NO2 = moles N2O5 2.28

Nitric acid is produced from the catalytic oxidation of ammonia
Problem Type 3 gram ↔ mole Nitric acid is produced from the catalytic oxidation of ammonia NH3(g) + O2(g)  NO(g) + H2O(g) a. How many moles of NO were produced from 824g of NH3? 4NH3(g) + 5O2  4NO(g) + 6H2O(g) 824g ? moles Units match 824 g NH3 = moles NO 48.4

Steps Involved in Solving Mass-Mass Stoichiometry Problems
Balance the chemical equation correctly Using the molar mass of the given substance, convert the mass given to moles. Determine the molar ratio. Using the molar mass of the unknown substance, convert the moles just calculated to mass. Amount in grams of known substance Amount in moles of known substance Amount in moles of unknown substance Amount in grams of unknown substance mw molar ratio mw

When N2O5 is heated, it decomposes:
Problem Type 4 grams ↔ grams When N2O5 is heated, it decomposes: How many grams of N2O5 are needed to produce 75.0 grams of O2? 2N2O5(g)  4NO2(g) + O2(g) ? grams 75.0 g 75.0 g O2 = grams N2O5 506

Stoichiometry Problem- Type 4
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al + O2 Al2O3 a. What are the reactants? b. What are the products? c. What are the balanced coefficients?

Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2  2Al2O3 = 6.50 g Al 1 mol Al 2 mol Al2O3 g Al2O3 ? g Al2O3 26.98 g Al 4 mol Al 1 mol Al2O3 6.50 x 2 x ÷ ÷ 4 = 12.3 g Al2O3

Acetylene gas (C2H2) is produced by adding water to calcium carbide:
How many grams of acetylene are produced by adding water to 5.0 grams of CaC2? CaC2 + 2H2O  C2H2 + Ca(OH)2 Balanced? 5.0 g ? grams 5.0 g CaC2 = grams C2H2 2.0

How many moles of CaC2 are needed to react completely with 49.0 g H2O:
CaC2 + 2H20  C2H2(g) + Ca(OH)2 How many moles of CaC2 are needed to react 49.0 g of H20? CaC2 + 2H20  C2H2(g) + Ca(OH)2 ? moles 49.0g Units match 49 g H2O = moles CaC2 1.36

Stoichiometry Review Problem
Suppose you want to figure out how many grams of ammonia you can produce if you react 60.0 g of hydrogen gas with excess nitrogen. Remember the balanced chemical equation gives the mole relationship between reactants and products. N H NH3 1) Take the grams of hydrogen gas (recall that it is diatomic) and convert it to moles: 2) Convert the moles of hydrogen gas to moles of ammonia, using the coefficients in the balanced chemical equation: 3) Convert from moles of ammonia to grams of ammonia using the molecular mass and solve the equation:

Calculating the Percent Yield
The predicted amount of product – which we have been doing with our stoichiometry problems has been for 100% yield, or the theoretical yield of a reaction. The theoretical yield is the maximum amount of product that can be produces in a given reaction. When chemical reactions take place, they are almost never 100% complete. A reaction may not go to 100% due to not all the reactants becoming involved, impurities in the reactants, competing side reactions, loss of product due to filtering, or just not getting it all out of the vessel. The actual yield is how much product can be collected – and measured. The percent yield is a ratio given by: Percent Yield = Actual Yield x % Theoretical Yield

Calcium carbonate is decomposed by heating, as shown in the following equation:
1) What is the theoretical yield of this reaction if 24.8 g CaCO3 is heated to yield 13.1 g CaO? Handle like a typical stoichiometry problem: grams to grams Ignore the 13.1 g CaO to solve. This is the actual yield. 2) What is the percent yield?

Chlorobenzene is used in the production of many different chemicals, such as aspirin, dyes and disinfectants. One method of preparing chlorobenzene is to react benzene C6H6, with chlorine according to the following equation: C6H6 + Cl C6H5Cl + HCl 1) When 36.8 g of C6H6 react with excess of Cl2, the actual yield of C6H5Cl is 38.8g? What is the theoretical yield? Handle like a typical stoichiometry problem: grams to grams 2) What is the percent yield?

Standard Molar Volume 1 mole of a gas occupies 22.4 liters of volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro At STP (Standard Temperature and Pressure): 1 mole of a gas occupies 22.4 liters of volume

Gas Stoichiometry #1 If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H2(g) + N2(g)  2NH3(g) 3 moles H2 + 1 mole N2  2 moles NH3 3 liters H2 + 1 liter N2  2 liters NH3

Gas Stoichiometry #2 How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen? 3 H2(g) + N2(g)  2NH3(g) 12 L H2 2 L NH3 = L NH3 8.0 3 L H2

Gas Stoichiometry #3 How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 22.4 L O2 g KClO3 2 mol KClO3 1 mol O2 = L O2 13.7