According to the following balanced equation, 2 formula units of iron (iii) oxide (fe2o3) can be formed by reacting 4 atoms of iron (fe) with 3 molecules of oxygen gas (o2). if 12 atoms of iron are reacted with 6 molecules of oxygen gas, which is the limiting reactant and how many atoms or molecules will be left over? 4fe + 3o2 -> 2fe2o3

You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol O₂
Then you find the limiting reagent by finding how much product each given amount of reactant can make.  Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe₂O₃ 
(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃ 
since oxygen produces the leas amount of product, oxygen is the limiting reagent.  since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
 find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
 find the number of atoms by multiplying that by Avogadro's number.
(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
therefore 4 atoms of Fe will be left over after the reaction happens.

I hope this helps.

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.