Question

Answer 1

Bài 7.

\(n_{Al}=\dfrac{4,05}{27}=0,15mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

0,15  0,1125           0,075       ( mol )

\(V_{O_2}=0,1125.22,4=2,52l\)

\(m_{Al_2O_3}=0,075.102=7,65g\)

Bài 8.

\(n_{O_2}=\dfrac{6,72}{22,4}=0,2mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

4/15   0,2                  2/15      ( mol )

\(m_{Al}=\dfrac{4}{15}.27=7,2g\)

\(m_{Al_2O_3}=\dfrac{2}{15}.102=13,6g\)