Black Friday is Here! Start Your Numerade Subscription for 50% Off!Join Today

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 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 112
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 137
Problem 138
Problem 139
Problem 140

Problem 138

When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms:

$$

\mathrm{Zn}(s)+\mathrm{S}_{8}(s) \longrightarrow \mathrm{ZnS}(s)[\text { unbalanced }]

$$

Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 $\mathrm{g}$ of $\mathrm{Zn}$ reacts with 52.4 $\mathrm{g}$ of $\mathrm{S}_{8}, 104.4 \mathrm{g}$ of $\mathrm{ZnS}$ forms.

(a) What is the percent yield of ZnS?

(b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?

Answer

a) 84,21 percent

b) $\mathrm{m}(\mathrm{ZnO})=16,36 \mathrm{g}$

$\mathrm{m}\left(\mathrm{SO}_{2}\right)=35,94 \mathrm{g}$

You must be logged in to bookmark a video.

...and 1,000,000 more!

OR

Join 5 million HS and College students

## Discussion

## Video Transcript

given initial quantities, we can calculate the percent yield as well as the theoretical you. The first step is to be sure your equation is balance. So in this case you ate here and an eight here and we know the initial masses as well as the actual yield, it's sometimes helpful to start by converting two moles using the molar Mass. So we do that by taking our initial mass and dividing by the molar mass so that we have initially 1.27 moles of zinc and 0.204 moles. Oh, sulfur, and we see that we form one 0.7 moles of zinc silver. To find the limiting re agent. We take both of our initial reactions and see how much product should for so we do this by using the ratio from the balanced equation. So if I have 1.27 moles of zinc, Balanced Equation tells May that for every eight moles of sink, I should form eight moles of product or 1.27 rules zinc sulfide. If I have 0.204 moles of sulfur, I will form eight moles of product for every one mole, the silver or 1.6 three moles. I think sulfide, since 1.27 is less than 1.63 My living re agent is ink, so that means I should make 1.27 moles of zinc changing to grants using the molar mass of zinc sulphide, which is 97. I went for four. It should make 100 and 24.0 grams. This is called the Theoretical Yield. And so my percent yield is what I actually make, or 100 and 4.4 grams divided by my theoretical yield 1 20 1 24 grams times 100 or 84.2%. So my percent yield for this reaction is 84.2%. There are some additional side reactions that happened Does think can react with oxygen to make zinc oxide and sulfur can react with oxygen to make sulfur dioxide. And to predict how much of each of these products we need to look at, how much zinc remains based on how much we used to make the yield. So again. Initially, we had 1.27 malls of zinc and 0.204 Moles of Sulfur The change is how much it took to make my moles of product. And so if I had 1.7 bowls of zinc sulfide and my ratio is for every eight moles of product, I need eight moles of zinc. That means I would use up 1.7 So my remaining amount is 1.27 minus 1.7 or 0.20 Moles Inc. And similarly, my change in sulfur looks at how much zinc sulfide formed, knowing that for every eight moles of zinc sulfide, I need one mole of sulfur. So I've used up 0.13 moles. So for leaving me with the remaining amount of zero 0.74 moles. So these were the quantities that we can use to predict how much product away. If I have point two moles zinc, my balanced equation tells me I'll make two moles of zinc oxide for every two moles of zinc. And then I'm gonna multiply that by the more mass of zinc oxide which is 81 point 379 So this tells me that I should make 16.27 grams of zinc oxide as a side product. If I take my 0.74 Moles of sulfur and my balanced equation tells me for every one mole of sulfur we'll make one mole of sulfur dioxide, which Ingram's multiplying by the molar mass of 64 0.58 grams for every mole. I should make 4.74 grams of sulfur dioxide as a side product.

## Recommended Questions

Zinc sulfide reacts with oxygen according to the reaction:

2 ZnS(s) + 3 O2(g)-2 ZnO(s) + 2 SO2(g)

A reaction mixture initially contains 4.2 mol ZnS and 6.8 mol O2. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

Zinc sulfide reacts with oxygen according to the reaction:

$$2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)

$$

A reaction mixture initially contains 4.2 $\mathrm{mol} \mathrm{ZnS}$ and 6.8 $\mathrm{mol} \mathrm{O}_{2}$ .

Once the reaction has occurred as completely as possible, what amount

(in moles) of the excess reactant remains?

Zinc sulfide reacts with oxygen according to the reaction:

A reaction mixture initially contains 4.2 $\mathrm{mol}$ Zns and 6.8 $\mathrm{mol}$ $\mathrm{O}_{2} .$ Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

Zinc reacts with iodine in a synthesis reaction: $\mathrm{Zn}+\mathrm{I}_{2} \rightarrow \mathrm{Zn} \mathrm{l}_{2}$

a. Determine the theoretical yield if 1.912 mol of zinc is used.

b. Determine the percent yield if 515.6 $\mathrm{g}$ of product is recovered.

When $10.0 \mathrm{g}$ zinc and $8.0 \mathrm{g}$ sulfur are allowed to react, all the zinc is consumed, $14.9 \mathrm{g}$ zinc sulfide is produced, and the mass of unreacted sulfur remaining is

(a) $2.0 \mathrm{g}$

(b) $3.1 \mathrm{g}$

(c) $4.9 \mathrm{g}$

(d) impossible to predict from this information alone

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid $\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)$ to form zinc sulfate $\left(\mathrm{ZnSO}_{4}\right)$ and molecular hydrogen $\left(\mathrm{H}_{2}\right)$. (a) Write a balanced equation for the reaction.

(b) If $0.0764 \mathrm{g}$ of $\mathrm{H}_{2}$ is obtained from $3.86 \mathrm{g}$ of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in part (b)?

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid $\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)$ to form zinc sulfate $\left(\mathrm{ZnSO}_{4}\right)$ and molecular hydrogen $\left(\mathrm{H}_{2}\right) .$ (a) Write a balanced equation for the reaction. (b) If 0.0764 g of $\mathrm{H}_{2}$ is obtained from $3.86 \mathrm{g}$ of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)?

Zinc sulfide (ZnS) occurs in the zinc blende crystal structure.

(a) If 2.54 g of ZnS contains 1.70 g of Zn, what is the mass ratio of zinc to zinc sulfide?

(b) How many kilograms of Zn are in 3.82 kg of ZnS?

The reaction between zinc and hydrochloric acid is carried out as a source of hydrogen gas in the laboratory:

$$

\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)

$$

If $325 \mathrm{mL}$ of hydrogen gas is collected over water at $25^{\circ} \mathrm{C}$ at a total pressure of $748 \mathrm{mm}$ Hg, how many grams of $\mathrm{Zn}$ reacted?

Hydrogen gas is produced when zinc reacts with sulfuric acid:

$$\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)$$

If 159 $\mathrm{mL}$ of wet $\mathrm{H}_{2}$ is collected over water at $24^{\circ} \mathrm{C}$ and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (The vapor pressure of water is tabulated in Appendix B.)