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Topic: Stoichiometry with unknown reactant  (Read 6346 times)

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ClarkKent

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Stoichiometry with unknown reactant
« on: July 24, 2005, 09:21:51 PM »
Hello, I am working on a summer assignment for AP Chem and about 3/4's the way through I ran across a question I haven't been able to start (My honors Chem teacher in years past wasn't that good of a teacher so the only reference material I have is google).  
The Question I have deals with stoichiometry but it has an unknown reactant.  

The questions is "Element M reacts with oxygen to produce a pure sample of MO2. Find the atomic mass and the identity of M if 9.6g of oxygen reacts with 16.5g of M to produce 26.1g of MO2" There is a multiple choice of "a.12(C) b.32(S) c.28(Si) d.40(Ca) e.55 (Mn)'

Sorry if the question is relatively easy but like I said my old teacher didn't cover this type of stoichiometry (we only worked with things with known reactants so I knew the mole ratio and everything).  Thanks in advance.

I tried using factor labeling started with 9.6g O2*(1 mol O2/ 16.0gO2)*(1 mol MO2/ 1 mol O2)*(26.1gMO2/ 1 mol MO2) but then after I did that I realized I don't even know if I can assume the mole ratio and that I got 15.66 which isn't even an answer. So basicly I'm entirely lost.
« Last Edit: July 24, 2005, 09:39:14 PM by ClarkKent »

Offline Qazzian

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Re:Stoichiometry with unknown reactant
« Reply #1 on: July 24, 2005, 11:40:58 PM »
Ok, here's what I got for this.

You have 9.6g of O reacting, so you can take

9.6g/ (16 g/mol) = 0.6mol O

The reaction is:

M + 2O <--> MO2

So there's 2 moles of O per mole of product, so you have

0.6/2 = 0.3 mol product. There's a 1:1 ratio between product and M, so there's 0.3mol M

The mass of M is 16.5g, and you can find the molar mass by

16.5g/0.3 mol = 55g/mol, which is Mn.


you were on the right track, but you made the mistake of using the molar mass of O2 being 16, that's the molar mass of O.
« Last Edit: July 24, 2005, 11:41:29 PM by Qazzian »
Biochemistry student. Third Year. University of Waterloo. Canada.

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