Enroll in one of our FREE online STEM bootcamps. Join today and start acing your classes!View Bootcamps

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

Problem 108

During studies of the reaction in Sample Problem 3.20

$$

2 \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)

$$

a chemical engineer measured a less-than-expected yield of $\mathrm{N}_{2}$ and discovered that the following side reaction occurs:

$$

\mathrm{N}_{2} \mathrm{H}_{4}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 6 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)

$$

In one experiment, 10.0 g of NO formed when 100.0 g of each

reactant was used. What is the highest percent yield of $\mathrm{N}_{2}$ that can

be expected?

Answer

89.7%

You must be logged in to like a video.

You must be logged in to bookmark a video.

## Discussion

## Video Transcript

given an equation in which you're given amounts of reactive issue could predict the amount of product that will form. This is called the Theoretical Yield. So in this case, we have 100 grams of each reacted and the product that we're looking for is end to because we have to amounts of reactant. One of them is the limiting re agent, so Onley one will determine how much product is formed. So the way to find this is too. First, convert each quantity two moles when we do this using the Moeller math. And then we can change from moules of the reactant two moles of the product that we do this using the mole ratio from the balanced equation. So we calculate this for both reactant and whichever smaller is our theoretical yield. To find the molar mass of each substance, we use the periodic table and we add the atomic masses of each substance. Time to sub script. So for into H four, it'll be two times 14.7 plus four times 1.8 or the more mass his 32.0 for six grams. Similarly, the molar mass for end to 04 is found by multiplying two times 14.7 plus four times oxygen, which is 15.994 Just 92 went 01 grams. We also need the molar mass of our product and to which is two times 14 with 007 We're 28.14 grams. If I have 100 grams of end to four h four, my first step is to change two moles dividing by molar Mass. And then I can change two moles because I know that there are three moles of nitrogen for every two moles and two h four. And so this amount would produce 4.68 moles and to similarly 100 grams and 204 changing two moles by dividing by the molar mass multiplying by the mole ratio, which is three moles and two for every one mole and two of four I find this produces 3.26 morals. Since this is the smaller quantity, that's the theoretical yield. We can change it to grams by multiplying by the molar mass of nitrogen. So my theoretically yield is 91.34 grams to find out how much is actually produced. We have to consider this side reaction. This will tell us how much nitrogen is taken away from the reaction in this case. I know that I'm producing 10 grams of N O, and I want to find how many grams of nitrogen that's equivalent to. So the first step is to change that two moles using the molar Mass by adding up nitrogen and oxygen. That's 30 wait 006 grams, and then I can find a relationship between my moles of nitrogen monoxide and a compound that's found in the previous equation. So in this case, I'll use my nature Die nitrogen tetroxide. I can then go back to the first reaction and see that that relates to the mole ratio of nitrogen. There are three moles of end to for everyone more of end to 04 and then finally, I can convert that two grams of nitrogen, and so if 10 grams of nitrogen monoxide is reacted, it should use up 9.34 grams of nitrogen. So the actual yield is how much I should have produced or 91.34 minus how much was taken away by the side reaction. We're 82 grams. I can then find my percent you by dividing my actual by my theoretical times, 100 commit 89.8% yield.