π The Study-to-Win Winning Ticket number has been announced! Go to your Tickets dashboard to see if you won! πView Winning Ticket

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 132

Manganese is a key component of extremely hard steel. The element occurs naturally in many oxides. A 542.3 -g sample of a manganese oxide has an Mn/O ratio of 1.00$/ 1.42$ and con-

sists of braunite $\left(\mathrm{Mn}_{2} \mathrm{O}_{3}\right)$ and manganosite $(\mathrm{MnO}) .(\mathrm{a})$ How many grams of braunite and of manganosite are in the ore? (b) What is

the $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ ratio in the ore?

Answer

a)

$\mathrm{m}(\mathrm{MnO})=154,29 \mathrm{g}$

$\mathrm{m}\left(\mathrm{Mn}_{2} \mathrm{O}_{3}\right)=388,05 \mathrm{g}$

b)

$\mathrm{Mn}^{3+}: \mathrm{Mn}^{2+}=1,92: 1$

You must be logged in to bookmark a video.

...and 1,000,000 more!

OR

Access this video now and improve your grades, **guaranteed**. **Study and win $10,000**. Cancel anytime.

Annual

$6.99/mo

30% off monthly price

billed $83.88 every 12 months

Quarterly

$8.99/mo

10% off monthly price

billed $26.97 every 3 months

Monthly

$9.99/mo

Pay as you go!

Unlimited access to over a million video solutions created by professors, teachers and educators

A community of 3,000,000 students & 20,000 educators.

Over 300 hours of STEM and Test Prep bootcamps

Ask unlimited questions

Unlimited personalized quizzes to test your knowledge.

And so much more!!

Unlimited access to over a million video solutions created by professors, teachers and educators

A community of 3,000,000 students & 20,000 educators.

Over 300 hours of STEM and Test Prep bootcamps

Ask unlimited questions

Unlimited personalized quizzes to test your knowledge.

And so much more!!

## Discussion

## Video Transcript

in a mixture oven or that contains both mag unease and oxide that are two different ion compounds. We can use our concept of percent composition to determine how many grams of this mixture RMN Oh, and how many grams air em into 03 So we know the ratio of manganese oxygen is one to 1.42 So if we look at our total mass, we can determine how much of that is oxygen. Oxygen is 1.42 parts out of the hole, which is one plus 1.42 or there should be 318.21 grams oxygen. We can also find how much of this his mag unease mag unease is one part of the whole where one divided by one plus 1.42 and so there are 224.9 grams of mm. We then look at dividing up between the two compounds. We know that our two compounds M n O. And mn two 03 So there are a total of three amends and four oxygen's. If we consider m n o, we know that it has 1/3 of the Mangga knees because it has one out of the three and it has 1/4 of the oxygen because it has one of the four oxygen's total, or 154 0.2 grams of this mixture, his M N o. You can do the same for em. In 203 You see that there are 2/3 of the mag unease present, and 3/4 of the oxygen or this compound is 388 0.5 grams of the total mixture. Look at the ratio of the ions. M n 20 is made up of the MN two plus I am and M in 203 is made up of the mn three plus ion. So using our masses, we can find out how many moles of each eye on is present. First, we changed two moles by dividing by the molar Mass and then noticing that there's one ah of mn two plus ions for everyone. Moul of compound or there are about two moles of mn two plus ions and using our mass for him. In 203 we changed two moles by dividing by the molar Mass. And we see in this equation there are two moles of mn three plus for every mole of compound. So this gives us five moles of MN three plus, so our ratio is a 5 to 2 ratio.

## Recommended Questions

Manganese is derived from pyrolusite ore, an impure manganese dioxide. In the procedure used to analyze a pyrolusite ore for its $\mathrm{MnO}_{2}$ content, a $0.533 \mathrm{g}$ sample is treated with $1.651 \mathrm{g}$ oxalic acid $\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)$ in an acidic medium. Following this reaction, the excess oxalic acid is titrated with $0.1000 \mathrm{M} \mathrm{KMnO}_{4}, 30.06 \mathrm{mL}$ being required. What is the mass percent $\mathrm{MnO}_{2}$ in the Ore?

$$

\begin{aligned}

\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{MnO}_{2}+\mathrm{H}^{+} & \longrightarrow \\

\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} & \text { (not balanced) } \\

\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{MnO}_{4}^{-}+\mathrm{H}^{+} \longrightarrow & \longrightarrow \\

\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} & \text { (not balanced) }

\end{aligned}

$$

A major source of boron is the mineral kernite, $\mathrm{Na}_{2}\left[\mathrm{B}_{4} \mathrm{O}_{5}(\mathrm{OH})_{4}\right] \cdot 3 \mathrm{H}_{2} \mathrm{O}$ . Calculate how many grams of boron can be produced from $1.0 \times 10^{3} \mathrm{kg}$ of a kernite -bearing ore if the ore contains 0.98$\%$ kernite by mass and the process has a 65$\%$ yield.

The balanced equation for a reaction in the process of reducing iron ore to the metal is

$$

\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})

$$

(a) What is the maximum mass of iron, in grams, that can be obtained from $454 \mathrm{g}(1.00 \mathrm{lb})$ of iron(III) oxide?

(b) What mass of $\mathrm{CO}$ is required to react with $454 \mathrm{g}$ of $\mathrm{Fe}_{2} \mathrm{O}_{3} ?$

The balanced equation for the reduction of iron ore to the metal using CO is $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$

(a) What is the maximum mass of iron, in grams, that can be obtained from $454 \mathrm{g}(1.00 \mathrm{lb})$ of iron(III) Oxide?

(b) What mass of $\mathrm{CO}$ is required to react with $454 \mathrm{g}$ of $\mathrm{Fe}_{2} \mathrm{O}_{3} ?$

An iron ore sample weighing 0.9132 g is dissolved in $\mathrm{HCl}(\mathrm{aq}),$ and the iron is obtained as $\mathrm{Fe}^{2+}(\mathrm{aq}) .$ This solution is then titrated with $28.72 \mathrm{mL}$ of $0.05051 \mathrm{M}$ $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .$ What is the mass percent $\mathrm{Fe}$ in the ore sample? $6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow$

$6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}.$

Some catalytic converters in automobiles contain the manganese oxides $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}$a. Give the names of $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}$.

b. Calculate the percent manganese by mass in $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}$.c. Explain how $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}$ are consistent with the law of multiple proportions.

Some catalytic converters in automobiles contain two manganese oxides: $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}.$

a. What are the names of these compounds?

b. What is the manganese content of each (expressed as a percent by mass)?

c. Explain how $\mathrm{Mn}_{2} \mathrm{O}_{3}$ and $\mathrm{MnO}_{2}$ are consistent with the law of multiple proportions.

Iron oxide ores, commonly a mixture of FeO and $\mathrm{Fe}_{2} \mathrm{O}_{3},$ are given the general formula $\mathrm{Fe}_{3} \mathrm{O}_{4}$ . They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hy- drogen. Balance the following equations for these processes.

$$\begin{array}{l}{\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g)} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \rightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)}\end{array}$$

A 3.104 g sample of an oxide of manganese contains 1.142 grams of oxygen. Write a balanced chemical equation for the reaction that produces the compound from $\mathrm{Mn}(\mathrm{s})$ and $\mathrm{O}_{2}(\mathrm{g})$.

The iron content of iron ore can be determined by titration with a standard KMnO, solution. The iron ore is dissolved in HCl, and all the iron is reduced to $\mathrm{Fe}^{2+}$ ions. This solution is then titrated with $\mathrm{KMnO}_{4}$ solution, producing $\mathrm{Fe}^{3+}$ and $\mathrm{Mn}^{2+}$ ions in acidic solution. If it required 38.37 mL of $0.0198 M$ KMnO $_{4}$ to titrate a solution made from $0.6128 \mathrm{g}$ of iron ore, what is the mass percent of iron in the iron ore?