WEBVTT
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when you're given steps of reaction that add up to
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given overall reaction, we confined the reaction by first
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writing the balanced equations for each step. So,
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for example, if we have nitrogen and oxygen reacting
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, we know that we have end to because nitrogen
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is always a die atomic molecule and reactions plus O
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two combined to form nitrogen monoxide. Nitrogen monoxide means
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there's one nitrogen and one oxygen. When we balance
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this, we see that we needed to in front
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of the N O. This nitrogen monoxide that we
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formed reacts with more oxygen to make nitrogen dioxide.
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Nitrogen dioxide is an 02 because there are three oxygen's
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on this side and only two on this side.
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We need to place the two here and a two
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here for the coefficients to balance the equation. The
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third step involves the nitrogen dioxide that was formed previously
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reacting with water vapor, or H 20 gas form
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nitric acid, which is H and 03 a que
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plus more nitrogen monoxide. And when we balance this
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equation, we see that we need to have a
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to here and a three here for coefficients to balance
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the equation to find the overall equation will add these
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altogether. But before we can do that, we
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need to be sure that the intermediaries or substances that
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are formed in one equation are used up in the
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next equation. So we need to be sure we
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have the same number of nitrogen mon oxides and dioxides
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on both sides. So to do this we can
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multiply equations by a factor. So here I see
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that I have three and oh, twos. But
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I've only produced two and o twos here. So
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if I look for my lowest common multiple, I
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multiply this equation by three. In this equation by
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two. This now gives me six and a two
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and four nitric acid. It's and to nitrogen monoxide
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. And I have six and three and six.
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So now every nitrogen dioxide that I produce will be
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used up if we consider our nitrogen mon oxides.
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I have six here and a total of four on
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this side. So to have a total of six
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on the product side, I need to multiply by
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two giving me a two here and two here and
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a four here. Now I can add the equations
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together and the intermediaries should cancel out on the react
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. Inside, I have to end two plus 202
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plus six, I know plus three oh too plus
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six ano too plus two h 20 And on the
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product side, I have four I know plus six
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and no to plus four h and 03 plus two
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n o. Since I have six n o twos
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on both sides and I have a total of six
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and nose on both sides, I can now simplify
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the equation by Canadian combining like terms. So I
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have to end to gas plus five because there are
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two plus three oh to gas plus two h+20
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gas makes four h and 03 which is a quist
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. And this is the complete balanced equation. We
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can use this equation to make conversion from one substance
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to another. So if I'm given a certain quantity
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of nitrogen in metric tons, I can find out
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how much nitric acid should be produced in tons as
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well. We'll need to change from metric tons two
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kilograms and the relationship is that everyone ton is 1000
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kilograms. Once I have kilograms, I can change
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two grams cause there are 1000 grams and every one
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kilogram. Once I have grams, I changed two
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moles and I do this using the molar mass of
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nitrogen. The molar mass is found using the periodic
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table, and we look at the individual molar mass
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of nitrogen and multiply it by two because there are
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too. So it's 28.14 grams per mole. Once
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I have moles of nitrogen, I can convert two
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moles of nitric acid. Coefficient of the balanced equation
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allows us to do this so we use what's called
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the mole ratio, which tells us there are four
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moles of nitric acid for every two moles of end
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to Once I have moles, I can change two
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grams using the molar mass of nitric acid, which
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I find using the individual molar masses and adding them
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together and equal 63 went 012 grams per mole and
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then after I have grams, I can change two
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kilograms and then finally two tons. So if I
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start up with my first quantity change two kilograms,
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then I changed two grams by multiplying by 1000 again
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. Then I changed two moles and then I can
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change two moles of nitric acid. Then I could
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change two grams and then we'll change two kilograms and
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then finally two metric tons. And so, by
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multiplying all the numbers on top and dividing by all
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the numbers on the bottom, we find that this
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should create 6.7 times 10 to the third metric tons
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of nitric acid.