`n_(H_2)=\frac{3,36}{22,4}=0,15(mol)`

`n_(NaOH)=0,8.1=0,8(mol)`

`2Al+2NaOH+2H_2O->2NaAlO_2+3H_2`

`0,1`        `0,1`                                 `0,1`                          `0,15`

`Al_2O_3+2NaOH->2NaAlO_2+H_2O`

`0,35`                 `0,7`                  `0,7`

`a,`

`m=0,1.27+0,35.102=38,4(g)`

`b,`

`C_(MNaAlO_2)=\frac{0,8}{0,8}=1M`

`c,`

`2Al+6HCl->2AlCl_3+3H_2`

`0,1`           `0,3`

`Al_2O_3+6HCl->2AlCl_3+3H_2O`

`0,35`              `2,1`

`C_(MHCl)=\frac{2,4}{1}=2,4(l)`

Đáp án:

\(\begin{array}{l} a,\ m=38,4\ g.\\ \%m_{Al}=7,03125\%\\ \%m_{Al_2O_3}=92,96875\%\\ b,\ C_{M_{NaAlO_2}}=1\ M.\\ c,\ V_{\text{dd HCl}}=2,1\ lít.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:\\ 2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2↑\ (1)\\ Al_2O_3+2NaOH\to 2NaAlO_2+H_2O\ (2)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ Theo\ pt\ (1):\ n_{Al}=n_{NaOH}=\dfrac{2}{3}n_{H_2}=0,1\ mol.\\ n_{NaOH}=0,8\times 1=0,8\ mol.\\ \Rightarrow n_{NaOH}\text{(pt2)}=0,8-0,1=0,7\ mol.\\ Theo\ pt\ (2):\ n_{Al_2O_3}=\dfrac{1}{2}n_{NaOH}=0,35\ mol.\\ \Rightarrow m=m_{Al}+m_{Al_2O_3}=0,1\times 27+0,35\times 102=38,4\ g.\\ \%m_{Al}=\dfrac{0,1\times 27}{38,4}\times 100\%=7,03125\%\\ \%m_{Al_2O_3}=\dfrac{0,35\times 102}{38,4}\times 100\%=92,96875\%\\ b,\ ∑n_{NaAlO_2}=n_{Al}+2n_{Al_2O_3}=0,1+0,7=0,8\ mol.\\ \Rightarrow C_{M_{NaAlO_2}}=\dfrac{0,8}{0,8}=1\ M.\\ c,\ PTHH:\\ 2Al+6HCl\to 2AlCl_3+3H_2↑\ (3)\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\ (4)\\ ∑n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,3+2,1=2,4\ mol.\\ \Rightarrow V_{\text{dd HCl}}=\dfrac{2,1}{1}=2,1\ lít.\end{array}\)

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