Stoichiometry Practice

Stoichiometry Practice

Balancing Equations

The first step to solving a stoichiometry problem is to balance the given chemical equation.

Balancing the equation will give you a better idea of the relationship between the reactants and the products, and will give you a ratio to work from when figuring out your conversions later.

Understanding the Equation

Subscripts

Coefficients

Understanding the Equation

Coefficients apply to the whole element or molecule they're in front of.

4NaCl means 4 sodiums and 4 chlorines.

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When there are subscripts involved, you'll multiply the coefficient with the subscript.

3O₂ means 6 oxygens total.

Subscripts apply only to the element they are directly behind.

Al₂O₃ means 2 aluminums and 3 oxygens.

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If there are parenthesis involved, then the subscript applies to each element inside.

(OH)₃ means 3 oxygens and 3 hydrogens.

 Remember!

HCl + O₂ → 2H₂O + Cl₂

4HCl + O₂ → 2H₂O + Cl₂

4HCl + O₂ → 2H₂O + 2Cl₂

4HCl + O₂ → 2H₂O + 2Cl₂

Your goal is to balance the elements on both the reactant and product side.

Start by balancing one element, and go from there.

​Reactants

Products

​Balanced!

Oxygens balanced.

Hydrogens balanced.

Chlorines balanced.

Drag and Drop

Balance the following equation.

$$Fe+Br_2\ \rightarrow\ FeBr_3$$

(Drag & drop the correct coefficient in front of each element.)

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Fe + ​

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Br₂ → ​

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FeBr₃

Drag these tiles and drop them in the correct blank above

1

2

3

4

5

Drag and Drop

Balance the following equation.

$$H_2+N_2\ \rightarrow\ NH_3$$

(Drag and drop the correct coefficient in front of each element.)

​

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H₂ + ​

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N₂ → ​

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NH₃

Drag these tiles and drop them in the correct blank above

3

1

2

4

5

Drag and Drop

Balance the following equation.

$$Al+H_2O\ \rightarrow\ Al_2O_3+H_2$$

(Drag and drop the correct coefficient in front of each element.)

​

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Al + ​

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H₂O → ​

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Al₂O₃ + ​

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H₂

Drag these tiles and drop them in the correct blank above

2

3

1

4

5

Drag and Drop

Balance the following equation,

$$Al\left(OH\right)_3\ \rightarrow\ Al_2O_3\ +\ H_2O$$

(Drag and drop the correct coefficient in front of each element.)

​

Placeholder

Al(OH)₃ → ​

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Al₂O₃ + ​

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H₂O

Drag these tiles and drop them in the correct blank above

2

1

3

4

5

Conversions

Once your equation is balanced, you have a good starting point to solve problems involving unit conversions.

A question may ask how many grams of a reactant you need, or how many moles of product will be produced from a reaction. Your first step will be to figure out what kind of conversion(s) you'll need to do.

Converting particles → moles?

Divide by Avogadro's number.

Converting moles → particles?

Multiply by Avogadro's number.

Particles - Moles

Converting moles → moles?

Use the coefficient ratio from the balanced equation.

Moles - Moles

Converting moles → grams?

Multiply by the molar mass.

Converting grams → moles?

Divide by the molar mass.

Moles - Grams

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Remember!

Multiple Choice

Choose which type of conversion is needed.

How many moles of water are in 24.3g of water?

• particles to moles

• grams to moles

• moles to moles

• moles to grams

Multiple Choice

Choose which kind of conversion is needed.

How many grams of carbon are in 8 moles of carbon?

• particles to moles

• grams to moles

• moles to moles

• moles to grams

Conversions - Moles to Moles

Consider the balanced equation below.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Remember that the coefficients in front give you a ratio between each substance, and if you change one, the rest will change by the same factor.

For 1 mole of C₃H₈, you'll need 5 moles of O₂.

For 2 moles of C₃H₈, you'll need 10 moles of O₂.

For 0.5 moles of C₃H₈, you'll need 2.5 moles of O₂.

Conversions - Moles to Grams

Let's go back to our earlier equation.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

See that 5 moles of O₂ are needed. How many grams of O₂ is this?

First, you need to determine what the molar mass of O₂ is.

The molar mass of O is 15.99g. So, the molar mass of O₂: 15.99g × 2 = 31.98g.

Next, multiply the molar mass with the number of moles: 31.98g × 5 = 159.9g

The mass of 5 moles of O₂ is 159.9g.

Conversions - Grams to Moles

Let's go back to our earlier equation.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

If 792.2g of CO₂ are produced from this reaction, how many moles is that?

​First, find the molar mass of CO₂.

The molar mass of C is 12.01g, O is 15.99g. So, the molar mass of CO₂: 12.01 + (15.99 × 2) = 43.99g

Next, divide the problem's mass by the molar mass:  792.2g ∕ 43.99g = 18.0 mol

792.2g of CO₂ is equal to 18.0 moles of CO₂.

Conversions - All Together

Now that you're familiar with the different kinds of conversions you can use, it's time to bring it all together.

You'll often need to use more than one kind of conversion to find the answer you're looking for when solving stoichiometry problems. Figuring out which ones you'll need is just as important as knowing how to use them!

Conversions - All Together

Consider the balanced equation. How many grams of CO₂ are produced from the combustion of 220.0g of C₃H₈?

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First, find how many moles of C₃H₈ are in 220.0g. Grams → Moles

Next, use your mole ratio to find how many moles of CO₂ are produced. Moles → Moles

Finally, find how many grams of CO₂ are produced. Moles → Grams

Let's get started!

Find the molar mass of C₃H₈.

C=12.01g, H=1.01g

(12.01g × 3) + (1.01g × 8) = 44.11g

Find how many moles are in 220.0g of C₃H₈.

220.0g ∕ 44.11g = 4.99mols

Find how many moles of CO₂ are produced.

There should be a 1:3 ratio between C₃H₈ and CO₂.

Since there are 4.99mols C₃H₈, 1:3 becomes 4.99:14.97

4.99 × 3 = 14.97mols CO₂.

Find how many grams of CO₂ are in 14.97mols.

14.97mols × 44.11g = 660.32g

There are 660.32g of CO₂ produced from this reaction.

How many grams of CO₂ are produced from the combustion of 220.0g of C₃H₈?

Grams → Moles

Moles → Moles

Moles → Grams

Find the molar mass of Ca(OH)₂ and CaCl₂.

Ca=40.08g, O=15.99g, H=1.01g, Cl=35.45g

Ca(OH)₂ 40.08 + (15.99 × 2) + (1.01 × 2) = 74.08g

CaCl₂ 40.08 + (35.45 × 2) = 110.98g

Find how many moles are in 65.0g of Ca(OH)₂.

65.0g ∕ 74.08g = 0.88mols

Find how many moles of CO₂ are produced.

There should be a 1:1 ratio between Ca(OH)₂ and CaCl₂.

Since there are 0.88mols Ca(OH)₂, 1:1 becomes 0.88:0.88

Find how many grams of CaCl₂ are in 0.88mols.

0.88mols × 110.98g = 97.66g

There are 97.66g of CaCl₂ produced from this reaction.

How many grams of CaCl₂ are produced from 65.0g of Ca(OH)₂?

Grams → Moles

Moles → Moles

Moles → Grams

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The END.