Conversion Factors from a Chemical Equation

Consider the equation:

4 NH3 + 5 O2 → 4 NO + 6 H2O

The coefficients in a given chemical reaction gives us the conversion factors to get from the number of

particles of one substance to the number of particles of another substance.

The above reaction tells us that 5 molecules of O2 are needed to react with 4 molecules of ammonia.

Mathematically, this is represented as:

Now, consider the following problem:

How many O2 molecules are required to react with 308 NH3 molecules?

Likewise, the following problem can be solved using the inverse of our expression:

How many NH3 molecules are required to react with 219 O2 molecules?

Our reaction:

4 NH3 + 5 O2 → 4 NO + 6 H2Omay also be interpreted on the molar level.

Four moles of NH3 molecules react with five moles of O2 to produce four moles of NO molecules and six moles of H2O molecules.On the molar level our Per relationships are written as:

Also, the inverse of each conversion factor is valid.

If the number of moles of either reactant or product is known, there is a one step conversion to the moles of any other species.

For example, if 3.20 moles of NH3 react according to our equation

4 NH3 + 5 O2 → 4 NO + 6 H2O

how many moles of H2O will be produced?

Example

Ammonia is formed directly from its elements. How many moles of hydrogen are needed to produce 4.20 moles of ammonia?

N2 + 3 H2 → 2 NH3

Given: 4.20 mol NH3 Wanted: mol H2

Per: 3 mol H2 / 2 mol NH3

Mass Calculations

Before you can solve any stoichiometry problem (how much or how many), you must have the reaction equation and the conversion factors between moles and quantities of given and wanted substances.

Once this information is determined, the solution of stoichiometry problems usually falls into a three-step mass-to-mass path.

Mass of given → Moles of given → Moles of wanted → Mass of wanted

Example

How many moles of H2O will be produced in a heptane, C7H16, burning reaction that yields 115 grams of CO2?

Write the balanced equation.

C7H16 + 11 O2 → 7 CO2 + 8 H2O

Calculate the molar mass of the given and wanted species

Molar mass = 44.01 g CO2 / mol CO2

18.02 g H2O/mol H2O With this information the problem can be solved.

Percent Yield

Consider the following problem: How many grams of CO2 will be produced by burning 66.0 g C7H16?

C7H16 + 11 O2 → 7 CO2 + 8 H2O

The answer to our problem, 203 g of CO2, is a theoretical yield.

This is the amount of product formed by the complete conversion of the given amount of reactant to product.

Factors such as impure reagents, incomplete reactions, and side reactions cause the actual yield to be less than the theoretical yield.

If the actual yield and the theoretical yield are known, the percent yield can be calculated.

% yield = (actual yield / theoretical yield) x 100%

If, in the previous example only 182 grams of CO2 had been produced instead of the calculated 203 grams, the percent yield would be:

Example

Calculate the theoretical yield and percent yield of nitrogen oxide, if 49.2 grams of nitrogen dioxide yields 8.90 grams of nitrogen oxide in the reaction of nitrogen dioxide and water to form nitric acid and nitrogen oxide.

Example with % yield givenSodium nitrite is produced from sodium nitrate by a decomposition reaction. Solid sodium nitrate decomposes to solid sodium nitrite and molecular oxygen. How many grams of sodium nitrate must be used to produce an actual yield of 60.0 g of sodium nitrite if the percent yield is 76.3% ?

2 NaNO3 → 2 NaNO2 + O2

Given: 60.0 g NaNO2 (act.) Wanted: g NaNO3

Per: 69.00 g NaNO2 / mol NaNO2 ; 2 mol NaNO3 / 2 NaNO2

85.00 g NaNO3 / mol NaNO3 ; 76.3 g NaNO2 (act) / 100 g NaNO2 (theo)

Limiting Reagents

Consider the following reaction:

C + O2 → CO2

Suppose we put three carbon atoms and two oxygen molecules into a reaction vessel. How many carbon dioxide molecules will we form?

This reaction is one-to-one; therefore, the oxygen is completely used up and one carbon atom remains.

Oxygen is the limiting reagent and determines how many carbon dioxide molecules can be formed.

One atom of carbon remains and is termed the excess reactant.

Limiting Reactants: Smaller-Amount Method

36.03 grams of carbon combines with 64.00 grams of O2 until one is totally consumed in the reaction:

C(g) + O2(g) → CO2(g)

How many grams of CO2 result?

How many grams of which element remain unreacted?

Assume that each reactant is limiting.If carbon is the limiting reactant (totally consumed), how many grams of CO2 will be produced?

If oxygen is the limiting reactant, how many grams of CO2 will be produced?

There is enough C to produce 132.0 grams of CO2; however, there is enough O2 to produce only 88.02 grams of CO2.O2 is the limiting reactant and the reaction stops when all the O2 is consumed.The limiting reactant is always the reactant that yields the smaller amount of product.

To find the amount of excess reactant that remains, calculate how much of that reactant will be used by the entire amount of limiting reactant.

We started with 36.03 g C. Reaction with the limiting reactant consumed 24.02 g C.The amount of remaining C is:

36.03g C (initial) - 24.02g C (used) = 12.01g C (remaining)

Procedure How to Solve a Limiting-Reactant Problem

1. Calculate the amount of product that can be formed by the initial amount of each reactant.

a. The reactant that yields the smaller amount of product is the limiting reactant.

b. The smaller amount of product is the amount (of product) that will be formed when all of the limiting reactant is used up.

2. Calculate the amount of excess reactant that is used by the total amount of limiting reactant

3. Subtract from the amount of excess reactant present initially, the amount that is used by all of the limiting reactant. The difference is the amount of excess reactant that is left.

Problem:

A solution containing 1.63g of BaCl2 is added to a solution containing 2.40g of silver nitrate, AgNO3. Find the number of grams of silver chloride, AgCl that can precipitate. Also determine which reactant was in excess, as well as the number of grams over the amount required by the limiting reactant.

BaCl2 + 2 AgNO3 → Ba(NO3)2 + 2 AgCl

Energy

Recall that nearly all chemical changes involve an energy transfer, usually in the form of heat.

The SI unit for energy is the joule (J).1 J is defined as 1 kg m2 / s2

An older energy unit is the calorie (cal).Originally defined as the amount of energy needed to raise the temperature of one gram of water 1 °C.

In terms of joules:1 calorie = 4.184 joules

or1 kcal = 4.184 kJ

Example

In a reaction between NH3 and HCl, 912 calories are transferred. Express this amount in J and kJ.

Given: 912 calWanted: J, kJ

Per: 4.184 J / cal

Thermochemical Equations

The heat given off or absorbed in a chemical reaction is called heat of reaction or formally, enthalpy of reaction, ∆H.

When a reaction is exothermic (releases heat to its surroundings), the enthalpy goes down, and ∆H has a negative value.

When a reaction is endothermic (absorbs heat from its surroundings), the enthalpy increases, and ∆H has a positive value.

∆G= ∆H -T∆S

An equation that includes a change in energy is a

Thermochemical Equation.

There are two ways to write these equations:

Simply write ∆H to the right of the equation.

For an exothermic reaction (energy is released):

2 C2H6(g) + 7O2(g) → 4 CO2(g)+ 6 H2O(l) ∆H = - 2855 kJ

or

2 C2H6(g) + 7O2(g) → 4 CO2(g)+ 6 H2O(l) + 2855 kJ

In an endothermic reaction, energy must be added to the reactants.For these types of reactions, heat is a “reactant”.

2 NH3(g) + 92 kJ → N2(g) + 3 H2(g)

or

2 NH3(g) → N2(g) + 3 H2(g) ∆H = +92 kJ

When writing Thermochemical equations, state symbols must be used.The magnitude of ∆H depends on the state (gas, liquid, solid) of the reactants and products.

Consider the following problem:

One of the fuels sold as “bottled gas” is butane, C4H10. Calculate the energy that may be obtained by burning 1.50 kg of butane if ∆H = - 5.77 x 103 kJ for this reaction.

2 C4H10(g) + 13O2(g) → 8 CO2(g) + 10 H2O(l)

Solution:

2 C4H10(g) + 13O2(g) → 8 CO2(g) + 10 H2O(l) + 5.77x103 kJ

Consider the complete combustion of liquid pentane, C5H10. If the reaction consumes 4.12 grams of oxygen, how many grams of water are produced?

2 C5H10 + 15 O2 → 10 CO2 + 10 H2O

O2: 32.00 g/mol

H2O: 18.02 g/mol

Consider the following reaction:

4 NH3 + 5 O2 → 4 NO + 6 H2ONH3: 17.03 g/mol H2O: 18.02 g/mol O2: 32.00 g/mol NO: 30.01 g/mol

How many grams of NH3 can be oxidized by 268 g of O2?

If the reaction consumes 538 g of NH3, how many grams of H2O will be produced?

How many grams of NH3are required to produce 404 g of NO?

∆H = - 75.8 kJ for the reaction S(s) + CO2(g) → SO2(g) + 2 C(s). When 10.1 grams of sulfur reacts with excess carbon monoxide, how many kilojoules of energy are evolved or absorbed?

Consider the following reaction:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

When 4.14 grams of C4H10 reacts with excess oxygen gas, the reaction yields 11.0 gram of carbon dioxide. What is the percent yield of this reaction?

C4H10 58.12 g/mol O2 32.00 g/mol CO2 44.01 g/mol H2O 18.02 g/mol

How many grams of hydrogen would have to react to produce 71.9 kJ of energy from the reaction

2 H2(g) + O2(g) → 2 H2O(g) + 484 kJ