This is possibly the hardest section of this chapter.
So, before we move into the actual calculations, we should know what they are.
Up until this point, we realize balanced equations tell us what happen in the reaction. Today, we know that balanced equations only tell us what should/usually happen in the reaction, but sometimes, the necessary conditions are not present.
For example, maybe the pressure/temperature is different, or the molar concentration is not as great.
Therefore, it is necessary to add more of one reactant than the equation predicts, because it is not possible for every atom or molecule to come together.
In other words, one reactant will be in excess (some of it will be left over) and another reactant will be used up completely.
This will be the limiting reactant, because it limits how much of a product we can actually produce.
If a bike is made from 2 wheels and 1 seat, and we have 4 wheels and 5 seats, what is the number of bikes produced?
The answer: 2 bikes...because if we just look at seats, there are 5 bikes possible. However, there are not enough wheels to make 5 bikes. That is why the limiting reactant is the wheel, and the seat is the excess reactant.
Now...let's get calculating!
We have the reaction: AgNO3 + CaCl2 ---> AgCl + Ca(NO3)2
When 32.0 g of CaCl2 reacts with 54.2 g of AgNO3, how many grams of AgCl are formed?
First, we balance the equation:
2 AgNO3 + CaCl2 ---> 2 AgCl + Ca(NO3)2
Now we want to convert both the reactants to the desired product. The smaller amount of product calculated is our answer.
32.0 g CaCl2 X 1 mole CaCl2 X 2 mol AgCl X 143.4 g AgCl = 82.6 g AgCl
111.1 g 1 mol CaCl2 1 mol AgCl
54.2 g AgNO3 X 1 mole AgNO3 X 2 mol AgCl X 143.4 g AgCl = 45.7 g AgCl
169.9 g 2 mol AgNO3 1 mol AgCl
Clearly, 45.7 g is smaller, so only 45.7 g of AgCl will be produced from this reaction. We can also realize that AgNO3 is the limiting reactant, since there is CaCl2 left over.
Now, if the question was "How much of the excess reactant is left over?"
We can calculate it. First, we want to see which one is limiting and which one is excess. We have done that above.
Next, since we are looking for how much of the excess reactant is left over, we want to convert the limiting reactant to the excess reactant using stoichiometry calculations.
The answer we get means 54.2 g of AgNO3 equals to ____ g of CaCl2.
So..... 54.2 g AgNO3 X 1 mole AgNO3 X 1 mole CaCl2 X 111.1 g CaCl2 = 17.7 g CaCl2
169.9 g 2 moles AgNO3 1 mole CaCl2
This means that 54.2 g of AgNO3 is equal to 17.7 g of CaCl2.
To get the excess, we simply take the excess reactant and minus our answer above.
32.0 g CaCl2 - 17.7 g CaCl2 = 14.3 g CaCl2
There are 14.3 g of CaCl2 left over, after 32.0 g of CaCl2 reacts with 54.2 g of AgNO3.
Now, it's time for a video and some problems!!!!
http://www.ausetute.com.au/exceslim.html
http://www.chemteam.info/Stoichiometry/WS-limiting-reagent.html
http://www.softschools.com/quizzes/chemistry/stoichiometry_limiting_reactants/quiz1210.html
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