7.1: Quantifying Reactants and products

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28 Terms

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Overview

Information Given by Chemical Equations

Mole-Mole Relationship

Mass Calculations

<p>Information Given by Chemical Equations</p><p>Mole-Mole Relationship</p><p>Mass Calculations</p>
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Information Given by Chemical Equations

A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction.

The coefficients of a balanced equation give the relative numbers of molecules.

<p>A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction.</p><p>The coefficients of a balanced equation give the relative numbers of molecules.</p>
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Describe the reaction: C2H5OH + 3O2 ---> 2CO2 + 3H2O

C2H5OH + 3O2 ---> 2CO2 + 3H2O

The equation is balanced.

All atoms present in the reactions are accounted for in the products.

1 molecule of ethanol reacts with 3 molecules of oxygen to produce 2 molecules of carbon dioxide and 3 molecules of water.

1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

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A balanced equation can predict the moles of product that a given number of _____ of reactants will yield.

moles

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Describe the number of moles that will react and the number of moles that will be produced from the following: 2H2O(l) → 2H2(g) + O2(g)

2 mol of H2O yields 2 mol of H2 and 1 mol of O2.

4 mol of H2O yields 4 mol of H2 and 2 mol of O2.

<p>2 mol of H2O yields 2 mol of H2 and 1 mol of O2.</p><p>4 mol of H2O yields 4 mol of H2 and 2 mol of O2.</p>
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Mole Ratio

The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation.

<p>The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation.</p>
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Example

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Example

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Steps for Calculating the Masses of Reactants and Products in Chemical Reactions

Steps for Calculating the Masses of Reactants and Products in Chemical Reactions

1. Balance the equation for the reaction.

2. Convert the masses of reactants or products to moles.

3. Use the mole ratio to calculate the number of moles

of the desired reactant or product.

4. Convert from moles back to masses

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Stoichiometry

Stoichiometry is the process of using balanced chemical equation to determine the relative masses of reactants and products involved in a reaction.

<p>Stoichiometry is the process of using balanced chemical equation to determine the relative masses of reactants and products involved in a reaction.</p>
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Stoichiometry Examples

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Chemical Quantities Overview

The Concept of Limiting Reactants

Calculations Involving Limiting Reactant

Percent Yield

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Stoichiometric Mixture

contains the relative amounts of reactants that matches the numbers in the balanced equation.

Example: N2(g) + 3H2(g) 2NH3(g)

<p>contains the relative amounts of reactants that matches the numbers in the balanced equation.</p><p>Example: N2(g) + 3H2(g) 2NH3(g)</p>
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Limiting Reactant

Limiting reactant is the reactant that runs out first and thus limits the amounts of products that can form.

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What is the limiting reactant of the following equation?

N2(g) + 3H2(g) ---> 2NH3(g)

H2

<p>H2</p>
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What is the limiting reactant of the following equation?

CH4 + H2O ---> 3H2 + CO

H2O is used up first, leaving 2 unused CH4 molecules unreacted.

The amount of products that can form is limited by water.

Water is limiting reactant.

Methane is in excess.

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Steps for Solving Stoichiometry Problems with Limiting Reactants

1. Write and balance the equation of the reaction.

2. Convert the known masses of reactants to moles.

Using the numbers of moles of reactants and the appropriate mole ratios, determine which is limiting

3. Use the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of desired product.

5. Convert from moles of product to grams of product, using molar mass (if this is required by problem).

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Examples

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Percent Yield

An important indicator of the efficiency of a particular labratory or industrial reaction.

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Theoretical yeld

The theoretical yield is the maximum amount of a given product that can be formed when the limiting reactant is completely consumed.

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Actual yield

The actual yield(amount actually produced) of a reaction is usually less than the theoretical yield

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Percent yield

The percent yield is the actual amount of a given product as the percentage of the theoretical yield.

Formula

<p>The percent yield is the actual amount of a given product as the percentage of the theoretical yield.</p><p>Formula</p>
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Example: We determined that 8.33 g C should be produced. 7.23 g of C is what was actually made in the lab. What is the percent yield of the reaction?

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Example

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To calculate amounts of reactants or products, you need a balanced chemical equation, a periodic table, and one known quantity.

1. Balance the given chemical equation.

2. Find the appropriate mole ratio from the balanced equation (product-to-reactant, reactant-to-product, reactant-to-reactant).

3. Convert the given quantity from units (molecules, mass, volume) to moles.

4. If you are calculating products from reactants, you may need to determine which reactant may be limiting.

5. Use the given molar quantity and the mole ratio to find moles of the unknown quantity.

6. Convert the moles of the unknown quantity to the desired units (molecules, mass, volume).

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When aluminum oxidizes in air, it forms aluminum oxide (Al2O3):

4Al (s) + 3O2 (g) → 2Al2O3 (s)

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If a 54 g sheet of aluminum foil oxidizes completely in excess oxygen, how many grams of aluminum oxide will be formed?

First determine if the equation is balanced.

~ Yes it is

Find the appropriate mole ratio from the balanced equation.

~ 4 : 2, or 2 : 1

Determine the molar mass of Al

~ (Periodic Table Needed) 26.982 or 27 g/mol

Determine the number of moles of the reactant that you were given.

~ ( Get rid of gAl in equation) 54 gAL* 1molAL/ 27 gAl= 2 MolAl

Determine the number of moles of the product that you will get.

~ ( Get rid of molAl) 2 molAl* 1Mol Al2O3/ 2 molAl= 1molAl2O3

You have 1 mol of Al2O3

Convert moles of product to mass. The molar mass of Al2O3 equals 102 g/molAl2O3=102 g/mol.

~ (Get rid of molAl2O3) 1molAl2O3* 102g Al2O3/ 1 molAl2O3= 102g of Al2O3

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