Chapter 10. Chemical Calculations and Chemical Equations

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Transcription:

Chapter 10 Chemical Calculations and Chemical Equations

Chapter 10

Equation Stoichiometry Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. General Steps 1. If you are not given it, write and balance the chemical equation for the reaction.

Equation Stoichiometry 2. Start your dimensional analysis in the usual way. 3. If you are not given grams of substance 1, convert from the unit that you are given to grams. This may take one or more conversion factors. 4. Convert from grams of substance 1 to moles of substance 1.

Equation Stoichiometry 5. Convert from moles of substance 1 to moles of substance 2 using the coefficients from the balanced equation to create the molar ratio used as a conversion factor. 6. Convert from moles of substance 2 to grams of substance 2, using its molar mass. 7. If necessary, convert from grams of 2 to the desired unit for 2. This may take one or more conversion factors.

Equation Stoichiometry Steps

Equation Stoichiometry Shortcut

Questions to Ask When Designing a Process for Making a Substance How much of each reactant should be added to the reaction vessel? What level of purity is desired for the final product? If the product is mixed with other substances (such as excess reactants), how will this purity be achieved?

Limiting Component

Limiting Component (2)

Limiting Reactant The reactant that runs out first in a chemical reaction limits the amount of product that can form. This reactant is called the limiting reactant.

Why substance limiting? (1) To ensure that one or more reactants are converted to products most completely. Expense P 4 (s) + 5O 2 (g) P 4 O 10 (s) + excess O 2 (g) Importance SiO 2 (s) + 2C(s) Si(l) + 2CO(g) + excess O 2 (g)

Why substance limiting? (2) Concern for excess reactant that remains danger P 4 (s) + 5O 2 (g) P 4 O 10 (s) + excess O 2 (g) ease of separation SiO 2 (s) + 2C(s) Si(l) + 2CO(g) + excess O 2 (g)

Limiting Reactant Problems Tip-off - You are given two amounts of reactants in a chemical reaction, and you are asked to calculate the maximum amount of a product that can form from the combination of the reactants. General Steps 1. Do two separate calculations of the maximum amount of product that can form from each reactant. 2. The smaller of the two values calculated in the step above is your answer. It is the maximum amount of product that can be formed from the given amounts of reactants.

Percent Yield Actual Yield Percent Yield = x 100% Theoretical Yield Actual yield is measured. It is given in the problem. Theoretical yield is the maximum yield that you calculate.

Why not 100% Yield? Reversible reactions Side Reactions Slow Reactions Loss during separation/purification

Conversions to Moles

Molarity Converts between moles of solute and volume of solution

Equation Stoichiometry (1)

Equation Stoichiometry (2) Tip-off - The calculation calls for you to convert from amount of one substance to amount of another, both of which are involved in a chemical reaction. General Steps 1. If you are not given it, write and balance the chemical equation for the reaction. 2. Start your dimensional analysis in the usual way.

Equation Stoichiometry (3) 3. Convert from the units that you are given for substance 1 to moles of substance 1. For pure solids and liquids, this means converting mass to moles using the molar mass of the substance. Molarity can be used to convert from volume of solution to moles of solute.

Equation Stoichiometry (4) 4. Convert from grams of substance 1 to moles of substance 1. 5. Convert from moles of substance 2 to the desired units for substance 2. For pure solids and liquids, this means converting moles to mass using the molar mass of substance 2. Molarity can be used to convert from moles of solute to volume of solution. 6. Calculate your answer and report it with the correct significant figures (in scientific notation, if necessary) and unit.