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Mass to Mass Conversions. Mole to Mole Conversions are the CRUCIAL LINK Follow the same steps: Step 1: Balance the equation Step 2: Write down what you.

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Presentation on theme: "Mass to Mass Conversions. Mole to Mole Conversions are the CRUCIAL LINK Follow the same steps: Step 1: Balance the equation Step 2: Write down what you."— Presentation transcript:

1 Mass to Mass Conversions

2 Mole to Mole Conversions are the CRUCIAL LINK Follow the same steps: Step 1: Balance the equation Step 2: Write down what you know Step 3: Determine conversion factors needed Step 4 Solve The only difference is now we need more conversions!

3 Example Problem #1: How many grams of potassium chloride are produced if 25 grams of potassium chlorate decomposes? KClO 3  KCl + O 2 1. Balance equation 2. Write down what you know 3. Determine conversions needed 4. Solve

4 2KClO 3  2KCl + 3O 2 25 grams KClO 3 x Grams KClO 3 to moles of KClO 3 Moles of KClO 3 to moles of KCl Moles of KCl to grams of KCl 1 mole KClO 3 122.6 g KClO 3 x 2 moles KClO 3 2 moles KCl x 1 mole KCl 74.6 g KCl = 15 grams KCl

5 Problem #2: How many grams of hydrogen are necessary to react completely with 50.0 grams of nitrogen in the following reaction: N 2 + H 2  NH 3

6 N 2 + 3H 2  2NH 3 50.0 grams N 2 x 1 mole N 2 28.0 g N 2 x 1 mole N 2 3 moles H 2 x 1 mole H 2 2.0 g H 2 = 10.7 grams H 2

7 Problem #3: How many grams of silver chloride are produced from 5.0 grams of silver nitrate reacting with an excess of barium chloride? AgNO 3 + BaCl 2  AgCl + Ba(NO 3 ) 2

8 2AgNO 3 + BaCl 2  2AgCl + Ba(NO 3 ) 2 5.0 grams AgNO 3 x 1 mole AgNO 3 169.9 g AgNO 3 x 2 moles AgNO 3 2 moles AgCl x 1 mole AgCl 143.4 g AgCl = 4.2 grams AgCl

9 Particle to Particle Conversions

10 Again, same concept as Gram-to-Gram. The crucial link is still the mole-to-mole ratio to convert between substances

11 Example Problem #1: Given the following reaction equation, how many atoms of copper can be produced from 4.96 x 10 24 atoms of aluminum? ___CuO + ___Al → ___Cu + ___Al 2 O 3 323 4.96 x 10 24 atoms Al x _1 mole Al__ 6.02x10 23 atoms Al x _3 mol Cu_ 2 mol Al x 6.02x10 23 __atoms Cu__ 1 mol Cu = 7.44 x 10 24 atoms Cu

12 OR Just like when you convert FROM liters TO liters (x and ÷ by 22.4), when you convert FROM particles TO particles (x and ÷ by 6.02 x 10 23 ) you can use a particle to particle ratio: 4.96 x 10 24 atoms Al x 2 atoms Al _3 atoms Cu_ = 7.44 x 10 24 atoms Cu

13 Example Problem #2: If 7.2 x 10 22 molecules of F 2 react with CH 2 S in a reaction, how many molecules of HF will be produced? ___CH 2 S + ___F 2  ___CF 4 + ___HF + ___SF 6 26 7.2 x 10 22 molecules F 2 x _1 mole F 2 __ 6.02x10 23 molecules F 2 x _2 mol HF_ 6 mol F 2 x 6.02x10 23 _molecules HF_ 1 mol HF = 2.4 x 10 22 molecules HF

14 OR 7.2 x 10 22 molecules F 2 x 6 molecules F 2 _2 molecules HF_ = 2.4 x 10 22 molecules HF

15 Problem #3: When sodium and hydrochloric acid are combined, 1.5 x 10 23 molecules of hydrogen are produced. How much sodium would have had to react? ___Na + ___HCl → ___NaCl + ___H 2 222 3.0 x 10 23 atoms Na

16 Volume to Volume Conversions

17 Avogadro’s principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. A mole (6.02 x 10 23 particles) of any gas at standard temperature and pressure takes up exactly 22.4 liters of space. Because of this, we can now do mole to volume conversions. This only works with gases! Solids and liquids take up different volumes due to the size of their particles and the forces between the particles. Gases are special this way. We will discuss this further in the future.

18 The same concept from Mass to Mass conversions can be applied to Volume to Volume conversions. Follow the same steps as all stoichiometry problems: 1. Balance equation 2. Write down what you know 3. Determine conversions needed 4. Solve

19 Example Problem #1: If 20.0 liters of oxygen are consumed in the below reaction, how many liters of carbon dioxide are produced? __C 3 H 8 + __O 2  __CO 2 + __H 2 O What conversions do we need? Liters of oxygen  moles of oxygen Moles of oxygen  moles of carbon dioxide Moles of carbon dioxide  liters of carbon dioxide

20 C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 20.0 L O 2 x 1 mole O 2 22.4 liters O 2 x 5 moles O 2 3 moles CO 2 x 1 mole CO 2 22.4 liters CO 2 = 12.0 Liters CO 2

21 Problem #2: If 0.38 L of hydrogen reacts with chlorine gas, what volume of hydrogen chloride gas will be produced? 1. We need a balanced equation.

22 Follow the rest of the Steps

23 Look at just the conversions What can be cancelled??

24 SO…….. Exception: For this type of problem (L to L), you can use a volume to volume ratio.

25 Problem #3: What volume of ammonia would be produced if the reaction begins with five liters of N 2 ? N 2 + 3H 2  2NH 3 5.0 L N 2 x 1 Liter N 2 2 Liters NH 3 =10 Liters NH 3

26 Problem #4: Sodium bicarbonate can be used to extinguish a fire. It decomposes to give carbon dioxide gas, which smothers the fire. If a sample contains 4.0 g of NaHCO 3, What volume of gas is produced at STP?

27 We have to go from MASS to VOLUME 1. We need a balanced equation.

28 0.53 Liters CO 2


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