# Chemistry word problem:Manganese is a key component of extremely hard steel. The element occurs naturally in many oxides. A 542.3g sample of Manganese oxide has a Mn : O ratio of 1.0 : 1.42,...

Chemistry word problem:Manganese is a key component of extremely hard steel. The element occurs naturally in many oxides. A 542.3g sample of Manganese oxide has a Mn : O ratio of 1.0 : 1.42, consists of braunite (Mn_2 0_3); Manganosite (MnO) What masses of braunite; manganosite are in the ore? What is the ratio of Mn^+3 Mn^+2 in the ore?

cgrant2 | High School Teacher | (Level 1) Valedictorian

Posted on

This problem can be broken down like so:

1. We know that Magnesium (Mn) has an atomic mass of 54.9g and Oxygen has an atomic mass of 16g. Together, the total mass of Maganosite is about 70.9g.

2. Total mass of is 542.3g and the ratio is 1/1.42 so that means that for every 1 Mn there are 1.42 O's.

3. Braunite is Mn2O3 which means there are 2 moles of Mn per 3 moles of oxygen. The mass of braunite is about 157.8g since 54.9g x 2= 109.9g and 16 x 3= 48g.

Let x= the amount of grams you need of each compound

70.9g(x) + 157.8g(x) = 542.3g

Do the math and you should get x=2.4 so round the answer to 2.

70.9 x 2= 141.8g of Manganosite

157.8 x 2= 315.6 g of Braunite

What is the ratio of Mn+2/Mn+3?

This means bascially the ratio of maganosite to braunite.

The ratio of Mn+2/Mn+3 is 2:2

I hope this helps a little.

drstunner | Student | eNotes Newbie

Posted on

I believe the way to do the problem is as follows:

Let x = your total grams of Braunite (Mn2O3), and y = total grams of Manganosite (MnO). Your first equation is:

x + y = 542.3g (given by the problem).

You know the molar ratio of Mn:O is 1:1.42. In other words, for each mol of Mn you have 1.42 moles of O. If you multiply each by their molar weight, you know you have 54.938g of Mn for every (1.42 moles * 15.9994g = 22.719g of O. So your O by mass is 22.719g/(54.938+22.719) = 29.256%.

This means in your 542.3g sample, 29.256%, or 158.653g of it is oxygen. Now you want to use Stochiometry to figure out how many moles of oxygen are in Braunite and Manganosite.

You know x grams of Braunite (the g of Braunite you start with) makes x/157.874 moles of Braunite (since molar mass of Braunite is 157.874) and that each mol of Braunite has 3 mol of Oxygen. If you take that number of mols of oxygen, and multiply it by the molar mass of oxygen, you can find the number of grams of oxygen in the Braunite. You will find that the value is .30403x grams Oxygen.

Do the same for the Manganosite to find that you'll have .225544y grams of oxygen in the Manganosite.

You know the total oxygen in the Braunite + total oxygen in the Manganosite = total mass of oxygen which as we deduced earlier is 158.653. So you know .30403x + .225544y = 158.653.

Use this and the equation x + y = 542.3 and you can figure out the values for x and y. I believe you get X = 463g (g of Braunite) and y = 79g (g of MnO) give or take. If you convert this back to moles you can find the ratio of Mn3+ to Mn2+ ions, keeping in mind that each mol of Braunite has 2 moles of Mn3+ whereas each mole of MnO only has 1 mol of Mn2+.

drstunner | Student | eNotes Newbie

Posted on

I'm fairly certain cgrant2's answer is wrong, and I would recommend the poster to double-check to make sure no one is led astray.

A few things in the poster's answer have inconsistent logic. You say let x = the amount of grams you need for each compound. The # of grams of Braunite and Manganosite should be different, yet you treat them both as x (the same value).

You also come up a molar ratio of 2:2, since you got x = 2. However assuming you mean that you have 2 moles of Braunite per 2 moles of Manganosite, then that means you have a total of 6 moles of Mn (4 from 2 moles of Mn2O3 and 2 from 2 moles of MnO) and 8 moles of O (6 from 2 moles of Mn2O3 and 2 moles from MnO) which means your ratio of Mn:O is 6:8, which is not equal to 1:1.42.

Please do correct me if I'm wrong, but I don't think the trail of logic you leave here makes sense (at least to me).