4Al + 3O2---->2 Al2O3
a) Ta có
n\(_{Al}=\frac{32,4}{27}=1,2\left(mol\right)\)
n\(_{O2}=\frac{21,504}{22,4}=0,96\left(mol\right)\)
=> O2 dư
Theo pthh
n\(_{Al2O3}=\frac{1}{2}n_{Al}=0,15\left(mol\right)\)
m\(_{Al2O3}=0,15.102=15,3\left(g\right)\)
c) Theo pthh
n\(_{O2}=\frac{3}{4}n_{O2}=0,225\left(mol\right)\)
n\(_{O2}dư=0,96-0,225=0,735\left(mol\right)\)
m\(_{O2}dư=0,735.32=23,52\left(g\right)\)
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