Chapter 13 – Stoichiometry, Study notes of Stoichiometry

These mole ratios are used to solve problems such as how many moles of carbon dioxide, CO2, would be produced from 6.25 moles of oxygen gas? Solution: 6.25 ...

Typology: Study notes

2021/2022

Uploaded on 08/05/2022

jacqueline_nel
jacqueline_nel 🇧🇪

4.4

(242)

3.2K documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Clark, Smith (CC-BY-4.0) GCC CHM 130 Chapter 13: Stoichiometry page 1
Chapter 13 Stoichiometry
Stoichiometry (STOY-key-OM-etry) problems are based on quantitative relationships between the
different substances involved in a chemical reaction.
13.1 Mole Ratio
The coefficients in a balanced equation given the moles of each substance in that equation.
For the combination reaction of hydrogen gas and nitrogen gas to produce ammonia, the coefficients give us
valuable information about the reaction:
N2(g) + 3 H2(g) 2 NH3(g)
For every 1 molecule of nitrogen that reacts, it needs three molecules of hydrogen to react with it. Together,
they produce 2 molecules of ammonia, NH3.
We can also say for every 1 mole of N2 that reacts, 3 moles of H2 reacts with it
to produce 2 moles of NH3.
These are mole-to-mole relationships/ratios.
o Given a balanced equations; any two compounds can be compared using mole-to-mole
relationships or mole ratios.
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
The mole ratios would be:
(3 mol CO2
5 mol O2) and (1 mol C3H8
4 mol H2O) and (3 mol CO2
4 mol H2O) and
(5 mol O2
4 mol H2O) and (5 mol O2
1 mol C3H8) and (3 mol CO2
1 mol C3H8), etc.
These mole ratios are used to solve problems such as how many moles of carbon dioxide, CO2, would be
produced from 6.25 moles of oxygen gas?
Solution: 6.25 moles O2 (3 mol CO2
5 mol O2) = 3.75 moles CO2
+
+
YouTube Video: Solving Stoichiometry Problems by weiner7000
STOP at 7:25 until you have read through the next three sections.
pf3
pf4
pf5

Partial preview of the text

Download Chapter 13 – Stoichiometry and more Study notes Stoichiometry in PDF only on Docsity!

Chapter 13 – Stoichiometry

Stoichiometry (STOY-key-OM-etry) problems are based on quantitative relationships between the different substances involved in a chemical reaction.

13 .1 Mole Ratio

The coefficients in a balanced equation given the moles of each substance in that equation. For the combination reaction of hydrogen gas and nitrogen gas to produce ammonia, the coefficients give us valuable information about the reaction:

N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

For every 1 molecule of nitrogen that reacts, it needs three molecules of hydrogen to react with it. Together, they produce 2 molecules of ammonia, NH 3.

 We can also say for every 1 mole of N 2 that reacts, 3 moles of H 2 reacts with it to produce 2 moles of NH 3.

These are mole-to-mole relationships/ratios. o Given a balanced equations; any two compounds can be compared using mole-to-mole relationships or mole ratios.

C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g)

The mole ratios would be:

3 mol CO 2

5 mol O 2 )^ and^ (

1 mol C 3 H 8

4 mol H 2 O )^ and^ (

3 mol CO 2

4 mol H 2 O)^ and

5 mol O 2

4 mol H 2 O)^ and^ (^

5 mol O 2

1 mol C 3 H 8 )^ and^ (^

3 mol CO 2

1 mol C 3 H 8 ), etc.

These mole ratios are used to solve problems such as how many moles of carbon dioxide, CO 2 , would be produced from 6.25 moles of oxygen gas?

Solution: 6.25 moles O 2 (

3 mol CO 2

5 mol O 2 )^ = 3.75 moles CO^2

YouTube Video: Solving Stoichiometry Problems by weiner STOP at 7:25 until you have read through the next three sections.

13. 2 Mass-Mass Stoichiometry

Steps:

  1. Grams of given  moles of given (Use the MM of given as your conversion factor.)
  2. Moles of given  moles of unknown (Use mole ratios from balanced equation.)
  3. Moles unknown  grams unknown (Use the MM of unknown as your conversion factor.)

 Important to include units & formulas for all substances- units cancel except wanted units.

Example: Calculate the mass of H 2 required to react with 8.75 g of O 2 according to the following balanced equations: O 2 (g) + 2 H 2 (g)  2 H 2 O(g)

Answer: 8.75 g O 2 (

1 mol O 2

32.00 g O 2 )^ (

2 mol H 2

1 mol O 2 )^ (

2.02 g H 2

1 mol H 2 )^ =^ 1.10 g H^2

(In your calculator: 8.75 ÷ 32.00 × 2 × 2.02 =)

13. 3 Mass-Volume Stoichiometry

OR

Recall: Avogadro’s Molar Volume is 22.4 L/mol for a gas only at STP

Steps:

  1. If given grams, use MM as your conversion factor to get to moles of the given -If given volume, use molar volume to get to moles of the given
  2. Use mol ratios to convert from moles of given to moles of unknown
  3. If asked to find grams, use MM as your conversion factor to get to grams of the unknown -If asked to find volume, use molar volume to get to liters of the unknown

Example: How many liters of oxygen gas are needed to react with 0.234 grams of SO 2 gas at STP? 2 SO 2 (g) + O 2 (g)  (^) 3 (g)

Answer: 0.234 g SO 2 (

1 mol SO 2

64.07 g SO 2 )^ (^

1 mol O 2

2 mol SO 2 )^ (

22.4 L O 2

1 mol O 2 )^ =^ 0.0409 L O^2

(In your calculator: 0.234 ÷ 64.07 ÷ 2 × 22.4 =)

Molar

Mass

Mole-Mole

Ratio

Molar

Mass

Grams of Given

Moles of Given

Moles of Unknown

Grams of Unknown

Molar

Mass

Mole-Mole

Ratio

Molar

Volume gas @ STP

Grams of Given

Moles of Given

Moles of Unknown

Liters of Unknown

Molar

Mass

Mole-Mole

Ratio

Molar

Volume gas @ STP

Liters of Given

Moles of Given

Moles of Unknown

Mass of Unknown

CHAPTER 13 PRACTICE PROBLEMS

Example 1 : N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

A. How many moles of N 2 are needed to completely react with 6.75 moles of H 2.

B. How many moles of NH 3 form when 3.25 moles of N 2 react?

C. How many moles of H 2 are required to produce 4.50 moles of NH 3?

Example 2: Consider the following reaction to produce iron, Fe (s):

Fe 2 O 3 (s) + 3 CO (g)  2 Fe (s) + 3 CO 2 (g)

A. Calculate the mass of CO needed to react completely with 50.0 g of Fe 2 O 3.

B. Calculate the mass of iron produced when 125 g of CO reacts completely.

C. Calculate the mass of CO 2 produced when 75.0 g of iron is produced.

Example 3 : Calculate the volume (in liters) of oxygen gas required to react with 50.0 g of aluminum at STP.

4 Al (s) + 3 O 2 2 Al 2 O 3 (s)

Example 4 : An automobile airbag inflates when N 2 gas results from the explosive decomposition of sodium azide (NaN 3 ),

2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) Calculate the mass of NaN 3 required to produce 50.0 L of N 2 gas at STP.

Answers to Practice Problems

Example 1 A  

 

 2

2 2 3 molH

6.75molesH^1 molN 2.25 mol N 2

B 

 

 2

2 3 1 molN

3.25molesN^2 molNH 6.50 mol NH 3

C 

 

 3

3 2 2 molNH

4.50molesNH^3 molH 6.75 mol H 2

Example 2 A  

  

 

 

 

 

 1 mole CO

28.01gCO 1 moleFeO

3 moleCO 159.70gFeO

50.0gFeO^1 moleFeO 2 3 2 3

2 3 2 3 = 26.3 g CO

B 

  

  

  

  

  

 1 mole Fe

55.85gFe 3 moleCO

2 moleFe 28.01gCO

125 gCO^1 moleCO = 166 g Fe

C 

 

  

  

  

  

 2

2 2 1 mole CO

44.01gCO 2 moleFe

3 moleCO 55.85gFe

75.0gFe^1 moleFe = 88.7 g CO 2

Example 3  

 

 

 

 

 

 2

2 2 1 mole O

22.4LO 4 moleAl

3 moleO 2 gAl

50.0g Al^1 moleAl 6_._ 98

= 31.1 L O 2

Example 4  

 

 

 

 

 

 3

3 2

3 2

2 2 1 molNaN

65.02gNaN 3 molN

2 molNaN 22.4LN

50.0LN molN 96.8 g NaN 3