1

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Gram to Gram Conversions

2

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions

3

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl3

3.45 g Al Alg 27.0

Almol

We must always convert to moles.

Now use the molar ratio.

Almol 2 AlClmol 2 3

Now use the molar mass to convert to grams.

3

3

AlClmol AlClg 133.3

17.0

Units match

gram to gram conversions

4

5

Molarity

Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)

When working problems, it is a good idea to change M

into its units.

mL 1000moles

Litermoles M

6

7

A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

What type of problem(s) is

this?

Molarity followed by

dilution.

Solutions

8

A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

1st:= mol L

3.73 gg 133.4

mol200.0 x 10-3 L

0.140

2nd: M1V1 = M2V2

(0.140 M)(10.0 mL) = (? M)(100.0 mL)0.0140 M = M2

molar mass of AlCl3

dilution formula

final concentration

Solutions

9

10

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

First write a balancedEquation.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

11

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now, let’s get organized. Place numerical Information and

accompanying UNITS below each compound.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 MLmol

? mL

35.0 mL

mL 1000mol 0.125

Lmol 0.125

Since 1 L = 1000 mL, we can use this to save on the number of conversions

Our Goal

12

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now let’s get to work converting.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 MLmol

? mL

35.0 mL

mL1000mol 0.125

Lmol 0.125

= mL NaOH

H2SO4

35.0 mL H2SO4

0.125 mol 1000 mL H2SO4

NaOH2 mol1 mol H2SO4

1000 mL NaOH0.102 mol NaOH

85.8

Units Match

Solution Stoichiometry:

shortcut

13

14

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

1st write out a balanced chemical

equation

Solution Stoichiometry

15

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl2

0.40 M 47.1 mL0.75 M? mL

= mL HCl

Ba(OH)2

47.1 mL

2

2

Ba(OH)

Ba(OH)

mL 1000 0.75mol

1 mol Ba(OH)2

HCl2 mol

0.40 mol HCl

HCl1000 mL 176

Units match

Solution Stoichiometry

16

17

Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL

0.135 mol L

25.00 mL

? mol L

18

Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL

0.135 mol L

25.00 mL

? mol L

= mol Ba(OH)2

L Ba(OH)2

25.00 x 10-3 L Ba(OH)2

Units Already Match on Bottom!

HClmL 23.28

HCl

HCl

mL 1000 mol 0.135

HCl

Ba(OH)

mol 2mol l

2 0.0629

Units match on top!

19

20

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.

Solution Stochiometry Problem:

21

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

Ca(OH)2(aq) + HNO3(aq) H2O(l) + Ca(NO3)2(aq)2 248.0 mL 19.2 mL

0.385 ML

mol 0.385

= mol(Ca(OH)2)

L (Ca(OH)2)

19.2 mLHNO3

3

3

HNO

HNO

mL 1000mol0.385

3

2

HNO 2molCa(OH) 1mol

48.0 x 10-3L

? M

units match!

0.0770

Solution Stochiometry Problem:

22

23

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Two starting amounts?

Where do we start?

Hide

one

24

Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.

L

mol 0.264L 10x 455g 213

moleg 25.63-

After you have worked the

problem, click here to see

setup answer

Try this problem (then check your answer):

25