\(a) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ n_{O_2} = \dfrac{3}{4},n_{Al} = \dfrac{3}{4}.\dfrac{5,4}{27} = 0,15(mol)\\ V_{O_2} = 0,15.22,4 = 3,36(lít)\\ c)\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ V_{O_2} = 2V_{CH_4} = 2,24.2 = 4,48(lít)\)
\(a.\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.2.....0.15\)
\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(b.\)
\(n_{CH_4}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)
\(0.1.........0.2\)
\(V_{O_2}=0.2\cdot22.4=4.48\left(l\right)\)
a) nAl= 5,4/27=0,2(mol)
PTHH: 4 Al + 3 O2 -to-> 2 Al2O3
nO2= 3/4. 0,2= 0,15(mol)
->V(O2,đktc)=0,15 x 22,4=3,36(l)
b) nCH4=2,24/22,4=0,1(mol)
PTHH: CH4 + 2 O2 -to-> CO2 + 2 H2O
nO2=2. 0,1=0,2(mol)
=>V(O2,đktc)=0,2.22,4=4,48(l)
a) PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
b) PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(V_{O_2}=2V_{CH_4}=2\cdot2,24=4,48\left(l\right)\)
