Unit 5/Unit Activity
Chemical Reactions - Fireworks directions
Task 1
Part A
Can click on this http://www.compoundchem.com/2014/07/02/the-chemistry-of-gunpowder/ or click on
Part B
Role of each type of chemical found in Part A. What does each do?
And this part of the activity
doesn't make sense, and I
don't know what I'm
supposed to do here...?
Part C
http://www.compoundchem.com/2014/07/02/the-chemistry-of-gunpowder/
or
Part D
Use this site:
http://scifun.chem.wisc.edu/chemweek/fireworks/Fireworks2017.htm
Part E
Watch this: https://www.youtube.com/watch?v=XtHtAiUVb7M (Jump ahead to the 1:38 mark.)
Part F
Use this site again: http://scifun.chem.wisc.edu/chemweek/fireworks/Fireworks2017.htm
Part G
Use this site again: http://scifun.chem.wisc.edu/chemweek/fireworks/Fireworks2017.htm
Part H
OK, the math….
There are 280g of black powder. Now, going back to Part C, use those percentages. In other words, calculate how
many grams of the 280g is KNO3? Do the same for charcoal (carbon) and sulfur. Make sure the % are in decimal
form before multiplying.
Part H is asking for how much (in grams) O2, CO2, and SO2 is created. So, you’ll need to use the reactions in Part D
to find those 3 gases.
Example of how do this: Benjamin Moore Premium Interior latex paint is 20% CaCO 3. Calcium carbonate
can produce CO2 gas through this reaction: CaCO3(s)------> CaO(s) + CO2(g)
If given 100.0g of this paint, how many grams of CO2 gas is produced?
Step 1
100.0g x 20%
or
100.0g x 0.20 = 20.00 g of CaCO3
Step 2
Now, 20.00 g CaCO3 needs to be converted to moles.
20.00g ÷ 100.0g/mol = 0.2000 moles
Step 3
Now, from the equation above, use the ratio of CaCO3 to CO2 is 1:1.
So, 0.2000 moles of CaCO3 = 0.2000 moles of CO2.
Step 4
Convert the 0.2000 moles of CO2 multiplied by the molar mass of carbon dioxide (44g/mol).
0.2000 mol x 44g/mol = 8.8 grams
DONE!
Use this to help guide you through all 3 calculations to find oxygen, carbon dioxide and sulfur dioxide.
You shouldn’t use the KClO3 or KClO4 reactions from Part D in Part H.
Part I
Here, you need to calculate 2 things: limiting reactant and how many grams of gases from Part H can be
formed. The grams of each you will be calculating are called a theoretical yield.
O2 So, in the equation you used for finding O2, there was only one reactant. Ask yourself: What would
the limiting reactant then be?? As far as mass of oxygen, if you know what the limiting reactant is, do
you really need to find the mass of it, again? Maybe, you already know it…….
CO2 To produce carbon dioxide from carbon, then the oxygen produced in the 1 st equation has to be
used in this equation. Think about the mole ratios in this equation. Comparing the moles of carbon to
oxygen from Part H, the one with the lowest number of moles is the limiting reactant. Use the limiting
reactant of those two to calculate the grams of CO2 gas produced.
SO2 In the production of sulfur dioxide, there is not going to be enough oxygen available. Like oxygen
above, you’ve already discovered the limiting reactant. Therefor, you’ve already calculated the mass of
this back in Part H.
Part J
The masses given to you in the directions in this problem are what are called actual yields. The masses
you just calculated in Part I are theoretical yields or best case scenario yields. Here you are doing what
you learned in the last section – Percent Yield before this activity. Need % yield of all 3 gases here.
Part K
You’re on your own here.
Good luck!!!
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