Mid-Term Review

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Molarity

Molarity is a method of representing concentration.  In this case, it is the ratio of the moles of solute to the liter of solution. The abbreviation for molarity is a capital M (MUST be capitalized, "m" means molality).

Reminder- Solute is the substance that is dissolved or dispersed in a solution. Solvent is the substance that actively separates or disperses the solute. Water is the most common solvent.

There are 3 possible unknowns for simple problems dealing with molarity- M, g of solute, L of solution.

Example 1- What is the molarity of a solution made from completely dissolving 5.25 g of NaCl in 435 mL of water?

Example 2- How many grams of NaCl are needed to make 200.0 mL of 0.0150 M solution?

Example 3- How many mL of water are needed to make a 0.250 M solution using 10.00 g of NaCl?

Stoichiometry: Limiting Reactants

To put it in simplest terms, the limiting reactant is whatever runs out first. Once one of the essential components runs out, you can't make anymore of the final product.

Basic Example 1- A company builds little red wagons.  Each wagon must have a body, 4 wheels and 2 axels. If the company has 125 bodies, 400 wheels and 300 axels in stock, how many complete wagons can they produce?

While there are several ways of approaching this, we'll solve for the maximum number of complete wagons that can be made from the number of each component given.

While there are sufficient bodies and axels to make more wagons, once the company makes 100 wagons, it will run out of wheels. The wheels are the limiting reactant and therefore all other amounts are dependent on them.

Basic Example 1 con't- How many axels will the company have left over after all the wheels are used?

300 axels were available - 200 axels used = 100 axels left over

Now let's look at chemistry example-

Example 2 - 204.3 g of sodium hydroxide reacts with 79.4 g of aluminum chloride. How many grams of the base will be produced?

After 43.8 g of  aluminum hydroxide is produced, all of the aluminum chloride will be consumed (used up), so the reaction stops. Therefore the aluminum chloride is the limiting reactant (also called the limiting reagent). There will be sodium hydroxide left over- it is in excess.

Example 2 con't - What will be the mass of the excess reactant remaining?

To determine the amount remaining, we must know how much we started with (204.3 g NaOH) and how much will be used. To find how much is needed, we start with the LR (in this case, the 74.9 g of aluminum chloride).

Therefore, 204.3 g initially - 67.3 g needed = 137 g NaOH remaining