May 17, 2024, 09:29:55 PM
Forum Rules: Read This Before Posting


Topic: Burning Magnesium  (Read 1564 times)

0 Members and 1 Guest are viewing this topic.

Offline beadone

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
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.

at
http://lecturedemos.chem.umass.edu/chemReactions4_1.html
which also looks wrong....as 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

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27682
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links