March 4th

More complicated STOICHIOMETRY!

Are you serious?!

Okay, fine.

Recall from the previous day's work that it's possible, using mole ratios, to convert mole-mole problems. Now we are going to learn how to convert from moles to grams and moles to particles and all of that good stuff!

LET'S GET STARTED!

Okay, the first thing is to recall the "mole map" we drew while working with previous chapters. Only this time, you can add stoichiometry and mole ratios to the mix.

This map is all nicely laid out for you, but in a real test or quiz, you won't get a map, so it's best to memorize all of this stuff now.

To convert more complicated quantities, you need to follow a few steps.

1. Make sure you have a balanced equation. You'll need one for the mole ratios.
2. Draw a mental mole map in your mind to know a) what quantity you have and b) the steps you need to get to the quantity you want
3. Write out the steps, cancelling out like quantities.
4. Make sure you have your sig figs in order.

Now let's try a few sample problems.

Example One:

Propane gas reacts with oxygen to form water and carbon dioxide. If 7.38g of oxygen is used in the reaction, how many mols of carbon dioxide will be produced?

1. First, write out the equation you get from the word equation.

C₃H₈ + O₂ ---> H₂O + CO₂

2. Next, balance the equation.

C₃H₈ + 5O₂ ---> 4H₂O + 3CO₂

3. Now, draw a mental mole map.

7.38g O₂ x 1 mol O₂/ 32g O₂ x 3 mol CO₂/ 5 mol O₂

4. Reduce the equation.

7.38g O₂ x 1 mol O₂/ 32g O₂ x 3 mol CO₂/ 5 mol O₂

= 0.138375 mols CO₂ with sig figs = 0.138 mols CO₂

Example Two:

Iron metal is dipped into sulfuric acid to produce soluble iron (III) sulfate and hydrogen gas. If 33 grams of iron metal is used in the reaction, how many grams of iron (III) sulfate will be produced?

1. First, write out the equation you get from the word equation.

 H₂SO₄ + Fe ---> Fe₂(SO₄)₃ + H₂

2. Next, balance the equation.

 3H₂SO₄ + 2Fe ---> Fe₂(SO₄)₃ + 3H₂

3. Now, draw a mental mole map.

 33g Fe x 1 mol Fe / 55.8g Fe x 1 mol Fe₂(SO₄)₃/ 2 mol Fe x 399.9g Fe₂(SO₄)₃/ 1 mol Fe₂(SO₄)₃

4. Reduce the equation.

33g Fe x 1 mol Fe / 55.8g Fe x 1 mol Fe₂(SO₄)₃/ 2 mol Fe x 399.9g Fe₂(SO₄)₃/ 1 mol Fe₂(SO₄)₃

= 118.25g Fe₂(SO₄)₃   with sig figs = 120g Fe₂(SO₄)₃

Now, that wasn't so bad, was it? :)

Here's a link to if you want to do some further stoichiometry problems!

Stoichiometry Problems!

And here's a link for the answers:

Stoichiometry Problem Answers