1. STOICHIOMETRY TUTORIAL
Paul Gilletti

2. Instructions: This is a work along tutorial
Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps.
Get a pencil and paper, a periodic table and a calculator, and let’s get to work.

3. (1-2-3) General Approach For Problem Solving:
1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?)
2. Determine what is given and the UNITS.
3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

4. Use the molar mass to convert grams to moles.
Converting grams to moles.
Determine how many moles there are in 5.17 grams of Fe(C5H5)2.
Given
Goal
units match
5.17 g Fe(C5H5)2
= moles Fe(C5H5)2
0.0278
Use the molar mass to convert grams to moles.
Fe(C5H5)2
2 x 5 x =
2 x 5 x =
1 x =

5. coefficients give MOLAR RATIOS
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + Ba(OH)2  2H2O BaCl2 1 1
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

6. Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
Units match
4.3 mol N2O5
= moles NO2
8.6
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
4.3 mol N2O5
= mole O2
2.2

7. gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
Units match
210 g NO2
= moles N2O5
2.28
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
? grams
75.0 g
75.0 g O2
= grams N2O5
506

8. 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.

9. 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.

10. 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.