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Stoichiometry Practice

Stoichiometry Practice

Assessment

Presentation

Chemistry

9th - 12th Grade

Practice Problem

Medium

NGSS
HS-PS1-7, MS-PS1-5

Standards-aligned

Created by

Mariah McConnell

Used 29+ times

FREE Resource

15 Slides • 16 Questions

1

Stoichiometry Practice

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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.

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Understanding the Equation

Subscripts

Coefficients

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Understanding the Equation

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

4NaCl means 4 sodiums and 4 chlorines.

-

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.

-

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

(OH)3 means 3 oxygens and 3 hydrogens.

 Remember!

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Fill in the Blank

How many aluminums are in the following molecule?

Al(NO3)3Al\left(NO_3\right)_3

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Fill in the Blank

How many nitrogens are in the following molecule?

Al(NO3)3Al\left(NO_3\right)_3

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Fill in the Blank

How many oxygens are in the following molecule?

Al(NO3)3Al\left(NO_3\right)_3

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HCl + O22H2O + Cl2

4HCl + O22H2O + Cl2

4HCl + O22H2O + 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.

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Drag and Drop

Balance the following equation.

Fe+Br2  FeBr3Fe+Br_2\ \rightarrow\ FeBr_3

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

Fe + ​
Br2 → ​
FeBr3
Drag these tiles and drop them in the correct blank above
1
2
3
4
5

10

Drag and Drop

Balance the following equation.

H2+N2  NH3H_2+N_2\ \rightarrow\ NH_3

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

H2 + ​
N2 → ​
NH3
Drag these tiles and drop them in the correct blank above
3
1
2
4
5

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Drag and Drop

Balance the following equation.

Al+H2O  Al2O3+H2Al+H_2O\ \rightarrow\ Al_2O_3+H_2

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

Al + ​
H2O → ​
Al2O3 + ​
H2
Drag these tiles and drop them in the correct blank above
2
3
1
4
5

12

Drag and Drop

Balance the following equation,

Al(OH)3  Al2O3 + H2OAl\left(OH\right)_3\ \rightarrow\ Al_2O_3\ +\ H_2O

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

Al(OH)3 → ​
Al2O3 + ​
H2O
Drag these tiles and drop them in the correct blank above
2
1
3
4
5

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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.

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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

media

Remember!

15

Multiple Choice

Choose which type of conversion is needed.

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

1

particles to moles

2

grams to moles

3

moles to moles

4

moles to grams

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Multiple Choice

Choose which kind of conversion is needed.

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

1

particles to moles

2

grams to moles

3

moles to moles

4

moles to grams

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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.

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Fill in the Blank

Consider the balanced equation:

C5H12+8O2  5CO2+6H2OC_5H_{12}+8O_2\ \rightarrow\ 5CO_2+6H_2O

If 2 moles of C5H12 are used instead, how many moles of O2 will be required?

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Fill in the Blank

Consider the balanced equation.

C5H12+8O2  5CO2+6H2OC_5H_{12}+8O_2\ \rightarrow\ 5CO_2+6H_2O

If 4 moles of O2 are used instead, how many moles of H2O will be produced?

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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.

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Fill in the Blank

A reaction needs 2 moles of N2. How many grams of N2 is this?

(The molar mass of N is 14.01g.)

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Fill in the Blank

A reaction produces 4 moles of NH3. How many grams of NH3 is this?

(The molar mass of N=14.01g, H=1.01g.)

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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.

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Fill in the Blank

A reaction produced 550g of CaCl2. The molar mass of CaCl2 is 110.98g.

How many moles of CaCl2 were produced?

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Fill in the Blank

A reaction uses 255.2g of HCl. The molar mass of HCl is 36.46g.

How many moles of HCl were used?

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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!

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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!

28

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

29

Fill in the Blank

Consider the balanced equation

Ca(OH)2+2HCl  CaCl2+2H2OCa\left(OH\right)_2+2HCl\ \rightarrow\ CaCl_2+2H_2O .

How many grams of CaCl2 will be produced from 65.0g of Ca(OH)2 in this reaction?

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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

How'd you do?

31

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Stoichiometry Practice

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