1. First write a balanced equation.
Gram to Gram Conversions
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)
First write a balanced equation.

2. Now let’s get organized. Write the information below the substances.
Gram to Gram Conversions
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)
3.45 g
? grams
Now let’s get organized. Write the information below the substances.

3. gram to gram conversions
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)
3.45 g
? grams
Units match
3.45 g Al
= g AlCl3
17.0
Let’s work the problem.
We must always convert to moles.
Now use the molar ratio.
Now use the molar mass to convert to grams.

5. When working problems, it is a good idea to change M into its units.
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.

7. What type of problem(s) is this? Molarity followed by dilution.
Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form mL solution. A 10.0 mL portion of the solution is then used to prepare mL of solution. Determine the molarity of the final solution.
What type of problem(s) is this?
Molarity followed by dilution.

8. Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form mL solution. A 10.0 mL portion of the solution is then used to prepare mL of solution. Determine the molarity of the final solution.
1st:
3.73 g
= mol L
0.140
200.0 x 10-3 L
molar mass of AlCl3
dilution formula
M1V1 = M2V2
2nd:
(0.140 M)(10.0 mL) = (? M)(100.0 mL)
final concentration
M = M2

10. Solution Stoichiometry:
Determine how many mL of M NaOH solution are needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
First write a balanced
Equation.

11. this to save on the number of conversions
Solution Stoichiometry:
Determine how many mL of M NaOH solution is needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
0.102 M
35.0 mL
? mL
Our Goal
Since 1 L = 1000 mL, we can use
this to save on the number of conversions
Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound.

12. Now let’s get to work converting.
Solution Stoichiometry:
Determine how many mL of M NaOH solution is needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
0.102 M
35.0 mL
? mL
Units Match
shortcut
H2SO4
35.0 mL
H2SO4
0.125 mol
1000 mL
NaOH
2 mol
1 mol
H2SO4
1000 mL NaOH
0.102 mol NaOH
= mL NaOH
85.8
Now let’s get to work converting.

14. Solution Stoichiometry
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

15. Solution Stoichiometry
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 mL
0.75 M
? mL
Units match
HCl
2 mol
Ba(OH)2
47.1 mL
HCl
1000 mL
176
= mL HCl
0.40 mol
HCl
1 mol
Ba(OH)2

17. First write a balanced chemical reaction.
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution. If it took mL of M hydrochloric acid to neutralize 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)
23.28 mL
25.00 mL
0.135 mol L
? mol L
First write a balanced chemical reaction.

18. Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution. If it took mL of M hydrochloric acid to neutralize 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)
23.28 mL
25.00 mL
Units match on top!
0.135 mol L
? mol L
= mol Ba(OH)2
L Ba(OH)2
0.0629
25.00 x L
Ba(OH)2
Units Already Match on Bottom!

20. We must first write a balanced equation.
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
We must first write a balanced equation.

21. Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
Ca(OH)2(aq) HNO3(aq)  2 2
H2O(l)
+ Ca(NO3)2(aq)
48.0 mL
19.2 mL
? M
0.385 M
HNO3
19.2 mL =
mol(Ca(OH)2)
L (Ca(OH)2)
0.0770
48.0 x 10-3L
units match!

23. Two starting amounts? Where do we start?
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
Hide
one
Two starting amounts? Where do we start?

24. After you have worked the problem, click here to see
Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.
After you have worked the problem, click here to see
setup answer