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In this video, we're going to be covering a few different stoichiometry problems.
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So the first step in solving a stoichiometry problem is to make sure that you have a balanced chemical equation.
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When we look at the chemical equation at the top of this whiteboard, we notice that we don't have the same number of oxygen atoms on both sides.
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I see that i have hydrogen peroxide, which is h2, and i have h2.
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H2o, which has one oxygen, and i have an o2.
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So as it is right now, h2o2 can only give us an even number of oxygen because of the o2.
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Any number that we put in front, multiply it by two will give us an even number.
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So i'm going to go ahead and put a two in front of h2o to make sure i have an even number of oxygen on my right -hand side of the equation.
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This now gives me four hydrogen and four oxygen.
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So if i put a two in front of h2o2, i now have four hydrogen and four oxygen on the left side as well, and my equation is now balanced.
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Now we can move on to finding how many moles of hydrogen peroxide we have to react in order to produce 0 .834 moles of 0 .2.
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So in order to do this, we need to set up our dimensional analysis, and we're going to use a mole ratio.
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And the mole ratio is going to come from our balanced equation.
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This is why it's important to make sure your equation is balanced properly.
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I'm trying to get my answer in moles of h2o2.
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So i'm going to put that in my numerator, and i need to cancel out my moles of o2.
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So i'm putting those in the denominator.
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Now i need to figure out what numbers belong with each of these units.
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In order to do this, i'm going to look at the coefficients in my chemical equation, and i'm going to write them into the proportion below.
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So for every two moles of hydrogen peroxide that i have reacting, one mole of oxygen will be formed.
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Now i can simply multiply across the top and divide by the bottom, and i get an answer of 1 .6 ,7, moles h2o2 and this answer is given to the correct number of significant figures.
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For our second stoichiometry problem, we're given the following chemical equation.
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It's a combustion reaction.
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Combustion reactions are notoriously more difficult to balance than other reactions.
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So in order to solve this balancing of the chemical equation, i'm going to go ahead and write out an inventory of each element on each side.
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So i have carbon present and there's six.
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I have h present and on my left hand side there are 12.
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And i have oxygen present and on my left hand side there are eight.
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On the right hand side, i have one carbon, two hydrogen, and three oxygen.
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So the first thing that i want to do in order to balance this equation is make sure that i have the proper number of hydrogen.
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And this is also going to make sure that my oxygen is now an even number.
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So i now have 12 hydrogen on my right hand side.
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And i have eight oxygen.
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But my carbon is still not balanced.
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So now i'll go ahead and i'll put a coefficient in front of my carbon dioxide.
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To balance out my carbon, this is also going to change the number of oxygen i have.
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I now have 18 oxygen on the right hand side...