Transcript 676-PROBLEM#4 ANS 11-6-10

PROBLEM 4 OF LIMITING REAGENT SET.
4 Al + 3O2  2Al2O3
FIRST DETERMINE IF THERE IS A LIMITING REAGENT.
FIND MOLES OF ALUMINUM METAL
MOLES = MASS/GFM Al
MOLES = 2.5g Al/ 26.98g/mol
MOLES = 0.092661mol Al
FIND MOLES OF OXYGEN GAS.
MOLES = MASS/GFM O2
MOLES = 2.5g O2 / 32.00g/mol
MOLES = 0.0781mol O2
RATIO BOTH REACTANTS TO FIND LIMITING REAGENT.
O2
Al
3
4
0.75
=
=
0.07810 mol O2
0.09266 mol Al
.842
The actual ratio is larger than the
theoretical, therefore the actual ratio
fraction is too large, the denominator is
to small Al is limiting.
4 Al + 3O2  2Al2O3
STEP 1 Identify known (the substance that can be converted to moles), choose the correct
equation to convert to moles.
MOLES = MASS/GFM Al
MOLES = 2.5g Al/ 26.98g/mol
MOLES = 0.092661mol Al
STEP 2 RATIO known to objective,
= 4 = 0.092661 ,
Al2O3
2
X
Al
X = 0.0463. mol Al2O3
STEP 3 Convert Objective to required units.
MOLES = MASS/GFM Al2O3
.0463 mol= mass/ 101.96
% yield = experimental/theoretical
3.5 / 4.7236 =74%
Mass =4.7236 g Al2O3