Today’s class dove into examples right away. Therefore, we will do the same!
1. A 0.750 L aqueous solution contains 90.0g of ethanol, C2H5OH. Calculate the molar concentration of the solution.
90.0g C2H5OH (1 mol C2H5OH / 46.0 g C2H5OH) ( 1 / 0.750 L) = [2.61]
2. How many liters of 0.100M HCl would be required to react completely with 5.00 grams of Ca(OH)2 in Ca(OH)2 + 2HCl à CaCl2 + 2H2O?
5.00 grams Ca(OH)2 ( 1 mol Ca(OH)2 / 74.1 g Ca(OH)2 ) ( 2 mol HCl / 1 ) ( 1 L / 0.100 mol HCl)
=1.35 L HCl is required
3. Using the same equation, if I added 15.0 grams of Ca(OH)2 with 75.0ml of 0.500M HCl, how many grams of CaCl2 would I form?
Remember, since two reactants are given you should find the limiting reactant.
15.0 g Ca(OH)2 ( 1 mol / 74.1 g) = 0.202 mol Ca(OH)2 present
0.075L (0.500 mol HCl / 1 L HCl) (1 mol Ca(OH)2 / 2 mol HCl) = 0.01875 mol Ca(OH)2 needed
Therefore, HCl is our limiting reactant. Now we can proceed to find the product formed.
0.075L (0.500 mol HCl / 1 L HCl) (1 mol CaCl2 / 2 mol HCl) (111.1 g CaCl2 / 1 mol CaCl2)
= 2.08 g CaCl2
If you are able to solve the question we just did, then you will have no problem in this area. This is the most difficult question that could be asked at the moment. Just remember, find your limiting reactant and then proceed as usual.
Posted by Andrew.
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