# Calculations from Equations Chapter 9

## Presentation on theme: "Calculations from Equations Chapter 9"— Presentation transcript:

Calculations from Equations Chapter 9
Larry Emme Chemeketa Community College

A Short Review

6.02 x 1023 Particles 1 MOLE Mole Weight

1 mole = 6.02 x 1023 ions 1 mole = 6.02 x 1023 atoms 1 mole = 6.02 x 1023 formula units 1 mole = 6.02 x 1023 molecule

The equation is balanced.
For calculations of mole-mass-volume relationships. The chemical equation must be balanced. The number in front of a formula represents the number of moles of the reactant or product. The equation is balanced. Al + Fe2O3  Fe + Al2O3 2 2 mol 1 mol

Introduction to Stoichiometry: The Mole-Ratio Method

Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products. Examples of questions asked: How much product from a specific amount of reactant. How much reactant do you start with to get so much product. What if the reaction does not go to completion? What is the percent yield. What if the sample is impure? What is the percent purity.

Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction.
The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

Examples

N2 + 3H2  2NH3 1 mol 2 mol 3 mol

N2 + 3H2  2NH3 1 mol 2 mol 3 mol

The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem. The mole ratio is used in the solution of every type of stoichiometry problem.

The Mole Ratio Method Convert the quantity of starting substance to moles (if it is not already moles) Convert the moles of starting substance to moles of desired substance. Convert the moles of desired substance to the units specified in the problem.

Step 1 Determine the number of moles of starting substance.
Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already done.

Step 2 Determine the mole ratio of the desired substance to the starting substance.
The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio.

In the following reaction how many moles of PbCl2 are formed if 5
In the following reaction how many moles of PbCl2 are formed if moles of NaCl react? 2NaCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2NaNO3(aq) Mole Ratio

Step 3. Calculate the desired substance in the units specified in the problem.
If the answer is to be in moles, the calculation is complete If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.

Step 3. Calculate the desired substance in the units specified in the problem.

Step 3. Calculate the desired substance in the units specified in the problem.

Step 3. Calculate the desired substance in the units specified in the problem.

Mole-Mole Calculations

Phosphoric Acid Phosphoric acid (H3PO4) is one of the most widely produced industrial chemicals in the world. Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4). Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4

Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4). Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 Moles starting substance: mol H2SO4 Step 2 The conversion needed is moles H2SO4  moles H3PO4 Mole Ratio

Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3F react. Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 The starting substance is 10.0 mol Ca5(PO4)3F Step 2 The conversion needed is moles Ca5(PO4)3F  moles H2SO4 Mole Ratio

Mole-Mass Calculations

The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction. If the mass of the starting substance is given, we need to convert it to moles.

We use the mole ratio to convert moles of starting substance to moles of desired substance.
We can then change moles of desired substance to mass of desired substance if called for by the problem.

Examples

Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4 Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4 grams H3PO4  moles H3PO4  moles H2SO4 The conversion needed is Mole Ratio

The conversion needed is
Calculate the number of grams of H2 required to form 12.0 moles of NH3. N2 + 3H2  2NH3 The conversion needed is moles NH3  moles H2  grams H2 Mole Ratio

Mass-Mass Calculations

Solving mass-mass stoichiometry problems requires all the steps of the mole-ratio method.
The mass of starting substance is converted to moles. The mole ratio is then used to determine moles of desired substance. The moles of desired substance are converted to mass of desired substance.

Diagram to Successful Calculations
mole wt of A moles of A Chemical equation moles of B grams of A mole wt of B grams of B

Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3 grams H2  moles H2  moles NH3  grams NH3

Limiting-Reactant and Yield Calculations

Limiting Reagent

The limiting reagent is one of the reactants in a chemical reaction.
It is called the limiting reagent because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reagent limits the amount of product that can be formed.

How many bicycles can be assembled from the parts shown?
From three pedal assemblies three bikes can be constructed. From eight wheels four bikes can be constructed. From four frames four bikes can be constructed. The limiting part is the number of pedal assemblies.

2 Wheels + 1 Frame + 1 Pedal→ 1 Bike

Steps Used to Determine the Limiting Reagent

Calculate the amount of product (moles or grams, as needed) formed from each reactant.
Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reagent; the other reactant is in excess. Calculate the amount of the other reactant required to react with the limiting reagent, then subtract this amount from the starting quantity of the reactant. This gives the amount of the substance that remains unreacted.

Examples

How many moles of HCl can be produced by reacting 4. 0 mol H2 and 3
How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reagent? H2 + Cl2 → 2HCl Step 1 Calculate the moles of HCl that can form from each reactant. Step 2 Determine the limiting reagent. The limiting reagent is Cl2 because it produces less HCl than H2.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many grams of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 1 Calculate the grams of AgBr that can form from each reactant. The conversion needed is g reactant → mol reactant → mol AgBr → g AgBr

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 2 Determine the limiting reagent. The limiting reactant is MgBr2 because it forms less AgBr.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
How many grams of the excess reactant (AgNO3) remain unreacted? MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50.0 g of MgBr2. The conversion needed is g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3 The amount of AgNO3 that remains is 100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3

Percent Yield

The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.

The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant.

The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion needed is g MgBr2 → mol MgBr2 → mol AgBr → g AgBr

MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction: MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq) Step 2 Calculate the percent yield. must have same units

Percent Purity

The percent purity of a reaction is the ratio of the theoretical yield (calculated) to the sample weight multiplied by 100.

g CO2 → mol CO2 → mol CaCO3 → g CaCO3
A 5.00 gram sample of impure CaCO3 (limestone) produces 1.89 grams of CO2 when heated. Calculate the percent purity of CaCO3 according to: CaCO3 (s)  CaO (s) + CO2 Step 1 Determine the actual grams (theoretical ) of CaCO3 in the original sample. The conversion needed is g CO2 → mol CO2 → mol CaCO3 → g CaCO3

A 5. 00 gram sample of CaCO3 produces 1. 89 grams of CO2 when heated
A 5.00 gram sample of CaCO3 produces 1.89 grams of CO2 when heated. Calculate the percent purity according to: CaCO3 (s)  CaO (s) + CO2 Step 2 Calculate the percent purity.

3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
484 gram of copper ore is oxidized by HNO3 to yield grams of Cu(NO3)2. Find the percent purity of copper in the ore according to: 3 Cu(s) + 8 HNO3(aq)  3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) The conversion needed is g Cu(NO3)2 → mol Cu(NO3)2 → mol Cu → g Cu

The End