Enroll in one of our FREE online STEM summer camps. Space is limited so join now!View Summer Courses

University of Maine

Problem 1
Problem 2
Problem 3
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84
Problem 85
Problem 86
Problem 87
Problem 88
Problem 89
Problem 90
Problem 91
Problem 92
Problem 93
Problem 94
Problem 95
Problem 96
Problem 97
Problem 98
Problem 99
Problem 100
Problem 101
Problem 102
Problem 103
Problem 104
Problem 105
Problem 106
Problem 107
Problem 108
Problem 109
Problem 110
Problem 111
Problem 113
Problem 114
Problem 115
Problem 116
Problem 117
Problem 118
Problem 119
Problem 120
Problem 121
Problem 122
Problem 123
Problem 124
Problem 125
Problem 126
Problem 127
Problem 128
Problem 129
Problem 130
Problem 131
Problem 132
Problem 133
Problem 134
Problem 135
Problem 136
Problem 138
Problem 139
Problem 140
Problem 212

Need more help? Fill out this quick form to get professional live tutoring.

Get live tutoring
Problem 128

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. N2(g) + O2(g) --> 2NO(g); 2NO(g) + O2(g) --> 2NO2(g); 3NO2(g) + H2O(g) --> 2HNO3(aq) + NO(g)

b. 2N2(g) + 5O2(g) + 2H2O(g) --> 4HNO3(aq)

c. 6.07x10^3 t

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## 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.

## Recommended Questions