em cần lời giải chi tiết

a)$ Al + 4HNO3 → Al(NO3)3 + NO + 2H2O $   (1)

$Al2O3 + 6HNO3 → 2Al(NO3)3 + 3H2O   $         (2)

$n_{NO} = 0,4 mol$

Theo (1): $n_{Al }= n_{NO} = 0,4 mol$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
m_{Al} \ =\ 0,4\ .\ 27\ =\ 10,8g\\
\%m_{Al} \ =\ \frac{10,8}{21,3} .\ 100\%\ =\ 50,7\%\\
\%m_{Al_{2} O_{3}} \ =\ 100\%\ -\ 50,7\%\ =\ 49,3\%\\
b) \ m_{Al_{2} O_{3}} \ =\ 21,3\ -\ 10,8\ =\ 10,5g\\
n_{Al_{2} O_{3}} \ =\ \frac{10,5}{102} \ mol\\
Theo\ ( 1) :\ n_{HNO_{3}} \ ( 1) =\ 4n_{Al} \ =\ 0,4\ .\ 4\ =\ 1,6\ mol\\
Theo\ ( 2) :\ n_{HNO_{3}}( 2) \ =\ 6n_{Al_{2} O_{3}} \ =\frac{10,5}{17} mol\\
n_{HNO_{3}} \ =\ n_{HNO_{3}} \ ( 1) \ +\ n_{HNO_{3}} \ ( 2) \ =\ 2,218\ mol\\
\rightarrow \ V\ _{HNO_{3}} \ =\frac{2,218}{2} \ =\ 1,109\ ( l)\\
c) \ n_{Al( OH)_{3}} \ =\frac{31,2}{78} =\ 0,4\ mol\\
Theo\ ( 1) :\ n_{Al( NO_{3})_{3}} \ ( 1) \ =\ n_{Al\ } =\ 0,4\ mol\\
Theo\ ( 2) :\ n_{Al( NO_{3})_{3}} \ =\ 2n_{Al_{2} O_{3}} \ =\ \frac{10,5}{51} mol\\
\rightarrow \ n_{Al( NO_{3})_{3}} \ =0,4\ +\ \frac{10,5}{51} =\ \frac{30,9}{51} \ mol\\
n_{Al^{3+}} \ =\ n_{Al( NO_{3})_{3}} \ =\frac{30,9}{51}  >n_{Al( OH) 3}\\
TH1:\ Al^{3+} \ còn\ dư:\\
Al^{3+} \ +\ 3OH^{-}\rightarrow \ Al( OH)_{3} \downarrow \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1,2 \ \ \leftarrow  \ \ \ \ 0,04\\
n_{NaOH} \ =\ n_{OH^{-}} \ =1,2\ mol\\
\rightarrow \ C_{M\ ( NaOH)} \ =\frac{1,2}{2} =\ 0,6\ ( l)
\end{array}$

TH2: Kết tủa đạt cực đại sau đó kết tủa bị hòa tan mất một phần:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
Al^{3+} \ +\ 3OH^{-} \ \rightarrow \ Al( OH)_{3} \downarrow \\
0,4 \ \ \leftarrow  \ \ 1,2 \ \ \leftarrow  \ \ \ \ 0,04\\
Al3+\ +\ 4OH^{-}\rightarrow \ AlO^{2-} \ +\ H_{2} O\\
\frac{10,5}{51} \ \ \ \ \ \ \ \ \rightarrow \frac{42}{51} \ \\
n_{NaOH} \ =\ n_{OH^{-}} \ =\ 2,024\ mol\\
V_{\ NaOH} \ =\frac{2,024}{2} \ =\ 1,012\ ( l)
\end{array}$