00:01
Hi, so to solve for this, we need to write first the balance equation for this reaction.
00:06
We have aluminum, oxygen, gas forming al2o3.
00:12
To balance this, we have to write 2 here.
00:14
That means we have 6 oxygen on the product side, so we'll write 3 here, and then 4 here.
00:19
This is the balance equation.
00:20
Now we need to determine the limiting reactant, convert the mass of reactants to moles.
00:26
We'll start with aluminum, 25 .7 grams of aluminum.
00:30
Convert to moles by dividing the molar mass of aluminum.
00:38
So we could cancel grams here, and we'll get 0 .370, sorry, not 0 .3, it's 953.
00:54
For oxygen, gas, we have 12 .1 grams, o2, divide the molar mass of o2, 32 grams per mole, cancel grams, and this is 0 .378.
01:10
Now to solve for the mole ratio, mole ratio is the number of moles of the reactant divided by the coefficient of the reactant from the balance equation.
01:20
So for aluminum, we have 0 .953 divided by the coefficient, which is 4...