Question

gấp 

Answer 1

a. Ta có PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)

\(n_{Al}=\dfrac{21,6}{27}=0,8mol\)

Vậy \(n_{Al_{Dư}}=\dfrac{36.15\%}{100\%}.27=0,2mol\)

Nên \(\%Al=100\%-15\%=85\%\)

\(\rightarrow n_{Al_2O_3}=\dfrac{85\%.36}{100\%}.102=0,3mol\)

Từ PT có: \(n_{O_2}=0,3.\dfrac{3}{2}=0,45mol\)

\(\rightarrow m_{O_2}=0,45.32=14,4g\)

b. Có: \(n_{Al}=0,3.\dfrac{4}{2}=0,6mol\)

\(\%m_{Al}=\dfrac{0,6.27}{21,6}.100=75\%\)