how many grams of Al2O3 be obtained if 13.5 g of aluminum completely reacts with oxygen as
4Al+3O2---->2Al2O3
Molar mass of Al is 27g/mol.
n=given mass/reaction mass=13.6/27=1/2
Now.,
n=given mole/reaction mole(Al2O3)=1/2/2=1/4
Now.,
n=mass/molar mass(Al2O3)
mass=1/4*102=25.50grams
4Al +3O2-------------> 2Al2O3
4Al----------->2Al2O3
1Al---------> 2/4 Al2O4
First convert grams to moles
Moles of aluminum = 13.5/27 =0.5
Now we have moles
0.5 moles of Al = 0.5 ×2/ 4 = p
4Al +3O2-------------> 2Al2O3
4Al----------->2Al2O3
1Al---------> 2/4 Al2O4
First convert grams to moles
Moles of aluminum = 13.5/27 =0.5
Now we have moles
0.5 moles of Al = 0.5 ×2/ 4 = 1/4
So we have moles of Al2O3 now but in question they asked us in grams. So we have to convert it to grams
Grams = moles x molar mass of Al2O3
Grams = 1/4 × 102 = 25.5
Well, let me calculate it in the most hilarious way possible! Buckle up!
So, first things first. We have 13.5 grams of aluminum. Now, if aliens invaded Earth and turned all our aluminum into tin foil hats, we would be pretty screwed, don't you think? But lucky for us, this aluminum is doing something much more useful - reacting with oxygen to form Al2O3!
Now, let's convert the grams of aluminum into moles. Since molar mass of Al is 27 g/mol, we divide 13.5 g by 27 g/mol, and guess what? We get 0.5 moles of aluminum. It's like finding half a donut in your pocket - unexpected but delicious!
Now, in that balanced equation, we see that for every 4 moles of aluminum, we get 2 moles of Al2O3. So, for 0.5 moles of aluminum, we can expect (drumroll, please) 0.25 moles of Al2O3. It's like getting a quarter, but instead of putting it in a piggy bank, you make fancy aluminum oxide!
Finally, we convert those moles into grams. The molar mass of Al2O3 is 102 g/mol. Multiply 0.25 moles by 102 g/mol, and voila! We have about 25.5 grams of Al2O3. It's as if we suddenly found 25.5 grams of gold under our pillows. I guess dreams do come true!
So, to sum it up, if 13.5 grams of aluminum completely react with oxygen according to the balanced equation, we would expect to obtain approximately 25.5 grams of Al2O3. That's a lot of shiny oxide, my friend!
To find the number of grams of Al2O3 obtained, you need to use stoichiometry. Stoichiometry is a branch of chemistry that deals with the ratios of reactants and products in a chemical reaction.
First, calculate the number of moles of aluminum (Al) present:
Number of moles of Al = Mass of Al / Molar mass of Al
= 13.5 g / 27 g/mol
= 0.5 mol
According to the balanced chemical equation, the stoichiometric ratio of aluminum (Al) to aluminum oxide (Al2O3) is 4:2, or 2:1. That means for every 2 moles of Al2O3, we need 2 moles of Al.
Since we have 0.5 moles of Al, we can now calculate the moles of Al2O3 produced:
Number of moles of Al2O3 = Number of moles of Al * (2 moles of Al2O3 / 2 moles of Al)
= 0.5 mol * (2/2)
= 0.5 mol
Finally, to find the mass of Al2O3, use the molar mass of Al2O3:
Mass of Al2O3 = Number of moles of Al2O3 * Molar mass of Al2O3
= 0.5 mol * (2 * 27 + 3 * 16) g/mol
= 0.5 mol * 102 g/mol
= 51 g
Therefore, if 13.5 g of aluminum completely reacts with oxygen, you would obtain 51 grams of aluminum oxide (Al2O3).
4Al+3O2---->2Al2O3
mols Al = grams Al/atomic mass Al.
Using the coefficients in the balanced equation, convert mols Al to mols Al2O3. You can see that 2 mol Al2O3 are formed for every 4 mols Al initially.
Then convert mols Al2O3 to grams. g = mols x molar mass = ?
Post your wor if you get stuck