Like

Report

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?

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 signed in to discuss.

Carleton College

Rice University

University of Kentucky

Numerade Educator

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.