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May 12th, 2018, 05:56 AM   #1
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determine mass of reactant not consumed

2Mg + O2 --> 2MgO
If 6.00g of magnesium and 2.00g of oxygen are available for reaction, identify and determine the mass of the reactant that is not consumed.

My attempted answer:

n(Mg)= (6.00g) / (24.305g/mol) = 0.247mol
n(O2)= (2.00g) / (32g/mol) = 0.0625mol

Mg:MgO
2:2
n(MgO)=0.247

O2:MgO
1:2
n(MgO)= 0.0625 x 2 = 0.125mol

Therefore, O2 is the limiting reagent and Mg is in excess.

0.247mol - 0.0625mol = 0.1845mol excess

m(Mg) = n x Mr
= 0.1845mol x 24.305g/mol
= 4.48g excess

The answer I got is 4.48g excess and thus this amount is not consumed.

I'm not sure whether this is correct. My teacher got around 1g and he said he doesn't know whether it's correct as well. I tried putting this question into a 'stoichiometry calculator' I found online, and it also gave a different answer.

I don't know who else to ask, so I hope someone on this forum would be able to help. =)

Last edited by skipjack; May 12th, 2018 at 05:44 PM.
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May 12th, 2018, 05:43 PM   #2
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The 0.0625mol of O$_2$ reacts with 0.125mol of Mg, which has mass 0.125*24.305g = 3.04g
so the mass of the remaining Mg is 6g - 3.04g = 2.96g.
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Last edited by skipjack; May 13th, 2018 at 03:51 AM. Reason: to correct the arithmetic.
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May 12th, 2018, 07:23 PM   #3
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Thanks skipjack,

I understand it now. The hard part now is to explain this solution to my teacher...I'm not sure why he doesn't know what the actual solution is even though he's been teaching chemistry for a long time...

when I calculate 0.122 x 24.305, I get 2.97g.

Last edited by skipjack; May 13th, 2018 at 03:54 AM.
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May 13th, 2018, 03:53 AM   #4
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I'd mistyped 24.305 when doing the calculation. I've amended my reply and avoided a slight rounding error, so the answer is now 2.96g.
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