A student combusted 0.500g of purified aluminum power with excess oxygen in an oxygen atmosphere according to the reaction, 4Al(s) + 3O2(g)= 2Al2O3(s)
A) What is the limiting reactant?
B) How many moles of Al were used?
C) How many moles of Al2O3 would form,
based on the moles of the limiting
reactant?
D) How many moles of O2 are required to
react completely with the aluminum?
E) How many grams of O2 are required to
react completely with the aluminum?
F) What is the theoretical yield of
AL2O3 in grams?
G) If the student collected 0.918 grams
of AL2O3 product, what was the
present yield of Al2O3 obtained?
2 answers
See your PbI2 problem above.
4al+3o2------------->2al2o3
Mass of al=2.7g
Moles of al=2.7÷27=0.1 moles
Comparing al and o2. According to balance equation
Al : o2
4 : 3
1 : 3÷4
0.1 moles : 0.75×0.1 moles
0.1 moles : 0.075 moles
So.
Mass of o2=moles×molar mass
=0.075×32
Mass=2.4g
Mass of al=2.7g
Moles of al=2.7÷27=0.1 moles
Comparing al and o2. According to balance equation
Al : o2
4 : 3
1 : 3÷4
0.1 moles : 0.75×0.1 moles
0.1 moles : 0.075 moles
So.
Mass of o2=moles×molar mass
=0.075×32
Mass=2.4g
1 answer