Hemet High Honors Chemistry

Hemet High  Honors Chemistry

9 Stoichiometry

Notes

Section 9.1 – Introduction to Stoichiometry

Types of Stoichiometry Problems

Given is in moles and unknown is in moles.
Given is in moles and unknown is in mass (grams).
Given is in mass and unknown is in moles.
Given is in mass and unknown is in mass.

Writing and Using Mole Ratios

Mole Ratio: a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles.
In chemical calculations, mole ratios are used to convert between:
N2(g) + 3H2(g)  2NH3 (g)
Three mole ratios can be derived from the balanced equation above:

Example: Li(s) + O2(g)  Li2O(s)

Section 9.2 – Ideal Stoichiometric Calculations

Mole-Mole Calculations

How many moles of ammonia (NH3) are produced when 0.60 mol of N2 reacts with H2.
Step 1: Write out equation and balance it
Step 2: Determine mole ratio of known (N2) to unknown (NH3).
Step 3: Multiply given amount of known by the correct mole ratio that will cancel out the known units, leaving you the unknown units

Example: How many moles of Chlorine are produced when 3.2 moles sodium chloride decomposes?

Mole-Mass Calculations

How many grams of aluminum oxide(Al2O3) are produced when 0.50mol Al reacts with O2?
Step 1: Write out equation and balance it
Step 2: Multiply by the mole ratio to get from known to unknown moles.
Step 3: Multiply the moles of unknown by the molar mass of the unknown to get grams.

Example: How many grams of Copper are needed to react with Sulfur to form 5.40 moles of copper sulfide?

Mass-Mole Calculations

Calculate the number of moles of H2O produced by the reaction of 5.40 g of O2 with an excess of H2. *excess means that there is enough to react.
Step 1: Write out equation and balance it.
Step 2: Multiply the given amount by the molar mass of the known to get moles.
Step 3: Multiply by the mole ratio to get from known to unknown moles.

Example: How many moles of hydrogen are needed to react with nitrogen to form 10.5 grams of ammonia (NH3)?

Example: How many moles of Lithium are formed by the decomposition of 3.40 grams of Lithium Oxide?

Mass-Mass Calculations

Calculate the number of grams of CO produced by the reaction of 3.25 g of C with an excess of O2.
Step 1: Write out equation and balance it.
Step 2: Multiply the given amount by the molar mass of the known to get moles.
Step 3: Multiply by the mole ratio to get from known to unknown moles.
Step 4: Multiply the moles of unknown by the molar mass of the unknown to get grams.

Example: How many grams of Oxygen are needed to make 20.0 grams of NO2? 2NO + O2 2NO2

Example: How many grams of SO3 are produced when 55.3 grams of SO2 reacts with an excess of O2.

Section 9.3 – Limiting Reactants and Percentage Yield

Limiting reactant: the substance that determines the amount of product that can be formed by a reaction.
Excess reactant: the substance that is not completely used up in a reaction.
Keep in mind that the reactant that is present in the smaller amount by mass or volume is not necessarily the limiting reactant.

Determining Limiting Reactant

What is the limiting reactant when 80.0g Cu reacts with 25.0g S? 2Cu(s) + S(s)  Cu2S(s)
Using the grams of reactant given, determine the moles of reactant. *Have
Using the balanced chemical equation, determine the mole ratio.
Multiply the moles of one reactant by the mole ratio to get the needed amount of the other reactant and compare to the amount you have.

*When you need more than you have, that is your limiting reactant.

*When you have more than you need, that is your excess reactant.

Example:

If 4.00 mol C2H4 is reacted with 9.50 mol O2, identify the limiting reactant and calculate the moles of excess reactant remaining. C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)

*Since we already have moles, we know what we have, we just have to find what we need.

Using a Limiting Reactant to Find Quantity of a Product

What is the maximum number of grams of Cu2S that can be formed when 45.0g Cu reacts with 24.0g S? 2Cu(s) + S(s)  Cu2S(s)
Find the limiting reactant and then use the moles of limiting reactant to calculate moles and then grams of Cu2S.

Limiting Reactant is ______, so we use the given amount of ______to find grams of Cu2S

Example

Calculate the moles of I2 produced when 80.0g I2O5 reacts with 28.0g CO.

I2O5(g) + 5CO(g)  5CO2(g) + I2(g)

Limiting Reactant is _____, so we use the given amount of _____ to find moles of I2. We already have converted to moles in the first part, so start with moles.

Percent Yield

Theoretical Yield: the maximum amount of product that could be formed from given amounts of reactants.
Actual Yield: the amount of product that actually forms when the reaction is carried out in the laboratory.
Percent Yield: the ratio of the actual yield to the theoretical yield expressed as a percent.

Percent Yield = actual yield x 100%units can be gram or mol, as

theoretical yieldlong as they are the same.

The percent yield is a measure of the of a reaction carried out in the laboratory.

Factors that can cause the actual yield to be less than the theoretical yield:

Calculating Theoretical Yield

What is the theoretical yield in grams of CaO if 24.8g CaCO3 is heated?

CaCO3(s)  CaO(s) + CO2(g)

Calculate the theoretical yield by converting the mass of the reactant to the mass of the product, using molar mass and mole ratios.

Example

What is the theoretical yield of AlCl3 when 3.00mol Al reacts with an excess of Cl2?

2Al + 3Cl2 2AlCl3

Calculating Percent Yield

What is the percent yield if 13.1g CaO is actually produced when 24.8g CaCO3 is heated? CaCO3(s)  CaO(s) + CO2(g)
Find the theoretical yield (13.9g CaO) and plug both the theoretical and actual yields into the percent yield equation.

Example

What is the percent yield if 2.85mol AlCl3 is actually produced when 3.00mol Al reacts with an excess of Cl2?

2Al + 3Cl2 2AlCl3

Another Way To Look At It

Think of percent yield as a grade on a test.
The Theoretical Yield is the number of points possible: 70 pts
The Actual Yield is the number of points you earned: 64 pts
Percent yield = 64/70 x 100% = 91%