a.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
4 3 2 ( mol )
0,2 0,1
\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,1.102=10,2g\)
b.\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2mol\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
4 3 2
0,2 < 0,1 ( mol )
0,1 1/15
\(m_{Al_2O_3}=n.M=\dfrac{1}{15}.102=6,8g\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH : 4Al + 3O2 -> 2Al2O3
0,2 0,1
\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
b. \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH : 4Al + 3O2 -> 2Al2O3
0,1 \(\dfrac{0,2}{3}\)
Xét tỉ lệ : \(\dfrac{0,2}{4}>\dfrac{0,1}{3}\) => Al dư , O2 đủ
\(m_{Al_2O_3}=\dfrac{0,2}{3}.102=6,8\left(g\right)\)
bài
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