can you calculate the mass(g) of AL2O3 that is produced when 2.5g of aluminum and 2.5g of oxygen react according to this equation : 4Al(s)+3O2(g)->2Al2O3(s)
im not understanding how to do this
mm. lasgania
This procedure will work almost all of the stoichiometry problems. This one is a limiting reagent problem (LR); you know that because amounts are given for BOTH rectants. I work these the long way.
1. Write and balance the equation. You have that. 4Al(s)+3O2(g)->2Al2O3(s)
2. Convert what you have into mols.
a. mols Al = grams/molar mass = ?
b. mols O2 = grams/molar mass = ?
3. Use the coefficients in the balanced equation to convert mols Al and mols O2 to mols Al2O3.
a. mols Al to mols Al2O3 is mols Al x (2 mol Al2O3/4 mols Al) = ?
b. Do the same for O2.
c. You see the mols Al2O3 are not equal from the two calculations which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.
4. Now convert mols to grams using the smaller LR reagent.
g = mols x molar mass = ?
You must read and follow step by step. I'll get you started. Following are my step by step instructions and I'll do the first few steps to show you how it's done.
1. Write and balance the equation. You have that. 4Al(s)+3O2(g)->2Al2O3(s)
2. Convert what you have into mols.
a. mols Al = grams/molar mass = ?
b. mols O2 = grams/molar mass = ?
2a. mols Al = grams/atomic mass = 2.5/27 = .0926 mols Al.
2b. mols O2 = grams/molar mass = 2.5/32 = 0.0781 mols O2
3. Use the coefficients in the balanced equation to convert mols Al and mols O2 to mols Al2O3.
a. mols Al to mols Al2O3 is mols Al x (2 mol Al2O3/4 mols Al) = ?
0.0926 mols Al x (2 mols Al2O3/4 mols Al) = 0.0926 x (2/4) = 0.0463 mols Al2O3 if we had 2.5 g Al and all of the O2 we needed.
b. Do the same for O2.
0.0781 mols O2 x (2 mols Al2O3/3 mols O2) = 0.0781 x (2/3) = 0.0521 mols Al2O3 if we had 2.5 g O2 and all of the Al we needed.
c. You see the mols Al2O3 are not equal from the two calculations which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.
Now you do step 3c and 4. Post your work if you get stuck.
4. Now convert mols to grams using the smaller LR reagent.
g = mols x molar mass = ?
the answer is mass Al2O3 = 101.96 g/mol x 0.04633 moles
= 4.72 g
could you help me
@DrBob222
how do i convert into mols?
2a and 2b above. mols = grams/molar mass
For Al that is grams/atomic mass = 2.5 g/ 27 = ?