a) PT :4 Al +3O2\(\underrightarrow{t^o}\) 2Al2O3
-nAl2O3(p/u)=\(\dfrac{2.295}{102}=0.0225\left(mol\right)\)
-Từ pt => nAl(p/u)=\(\dfrac{1}{2}\)nAl2O3(p/u)=0.01125(mol)
=>mAl(p/u)=0.01125.27=0.30375(g)
H%=\(\dfrac{m_{Al}\left(p/u\right)}{m_{Al}}.100=\dfrac{0.30375}{1.35}.100=22.5\%\)
b)
-Từ pt => -Từ pt => nO2(p/u)=\(\dfrac{3}{2}\)nAl2O3(p/u)=0.03375(mol)
=> VO2(p/u)=0.03375.24=0.81(l)
-Coi O2 chiếm 21% thể tích không khí =>
V kk cần : 0.81:21%=3.8571(l)
Vậy...