Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord! Numerade Educator ### Problem 128 Hard Difficulty # Various nitrogen oxides, as well as sulfur oxides, contribute to acidic rainfall through complex reaction sequences. Nitrogen and oxygen combine during the high-temperature combustion of fuels in air to form nitrogen monoxide gas, which reacts with more oxygen to form nitrogen dioxide gas. In contact with water vapor, nitrogen dioxide forms aqueous nitric acid and more nitrogen monoxide. (a) Write balanced equations for these reactions.(b) Use the equations to write one overall balanced equation that does not include nitrogen monoxide and nitrogen dioxide. ($(\text { ) How }$many metric tons (t) of nitric acid form when 1350 t of atmospheric nitrogen is consumed$(1 \mathrm{t}=1000 \mathrm{kg}) ?$### Answer ## a)$\mathrm{N}_{2(g)}+\mathrm{O}_{2(g)} \longrightarrow 2 \mathrm{NO}_{(g)}$$2 \mathrm{NO}_{(g)}+\mathrm{O}_{2(g)} \longrightarrow 2 \mathrm{NO}_{2(g)}$$3 \mathrm{NO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(g)} \longrightarrow \mathrm{NO}_{(g)}+2 \mathrm{HNO}_{3(a q)}$b)$2 \mathrm{N}_{2(g)}+5 \mathrm{O}_{2(g)}+2 \mathrm{H}_{2} \mathrm{O}_{(g)} \longrightarrow 4 \mathrm{HNO}_{3(a q)}$c)$\mathrm{m}\left(\mathrm{HNO}_{3}\right)=6,06^{*} 10^{9} \mathrm{g}=6059 \mathrm{t}\$

#### Topics

Chemical reactions and Stoichiometry

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### Video Transcript

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

University of Maine

#### Topics

Chemical reactions and Stoichiometry