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Topic: Burning Magnesium  (Read 1564 times)

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Offline beadone

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Burning Magnesium
« on: September 06, 2014, 12:17:38 AM »
I was helping my child with a question on the weight of oxygen required when .24 grams of magnesium was burnt in air and the resultant weight of mgO
It was straight forward in my view

2mg + 1O2 = 2mgO
so .24g = .01 moles of mg, therefore we need .01/2 moles of O2 (Molar mass of O2 is 32 grams)

so the final equation is .24 + 0.005x32 = .4grams . Our initial problem was that the worked result given gave and answer of 0.53grams which we think is wrong.
but I also looked up other similar equations on the net and found the following

2Mg(s) + O2(g) → 2MgO(s)

The stoichiometric factor is 2 moles of magnesium are burned for every 1 mole of oxygen (2mol Mg/1mol O2). If the magnesium strip weighs 1 gram, then there is 0.04 mol of magnesium (1 gram divided by 24.3 grams/mol Mg) available in the reaction. The amount of oxygen required to completely react with the magnesium strip is:

0.04 mol Mg x (1 mol O2 / 2 mol Mg) = 0.02 mol O2 x 16 g/mol O2 = 0.32 gram O2.

which also looks they use 16g/mol of o2 and forget to multiply by 2 or should use 32 as the molecular mass of oxygen not 16 which is the molecular weight. Now I am unsure of the
correct approach.

Offline Borek

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Re: Burning Magnesium
« Reply #1 on: September 06, 2014, 03:08:10 AM »
0.4 g MgO from 0.24 g of magnesium is a correct answer.

Please capitalize your formulas properly - it is Mg and MgO.
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