Đáp án:

a) $2Al + 3H_{2}SO_{4}→Al_{2}(SO_{4})_{3} + 3H_{2}↑$ (1)

b) Ta có: $n_{Al}=\dfrac{m}{M}=\dfrac{8,1}{27}= 0,3 (mol)$

Theo PT (1): $n_{H_{2}}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2} × 0,3= 0,45 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,45 × 22,4= 10,08 (l)$

c) Theo PT (1): $n_{H_{2}SO_{4}}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2} × 0,3= 0,45 (mol)$

⇒ $m_{H_{2}SO_{4}}=n × M=0,45 × 98= 44,1 (g)$

⇒ $\text{C%$H_{2}SO_{4}=\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{44,1}{230}$ × 100%= 19,2%}$

d) $Fe_{2}O_{3} + 3H_{2}\xrightarrow{t^o}2Fe + 3H_{2}O$ (2)

$n_{Fe_{2}O_{3}}=\dfrac{m}{M}=\dfrac{32}{160}= 0,2 (mol)$

Lập tỉ lệ: $\dfrac{0,2}{1}>\dfrac{0,45}{3}$

⇒ $Fe_{2}O_{3}$ dư, tính theo mol $H_{2}$

Theo PT (2): $n_{Fe_{2}O_{3}}=\dfrac{1}{3}n_{H_{2}}=\dfrac{1}{3} × 0,45= 0,15 (mol)$

⇒ $m_{Fe_{2}O_{3}(dư)}=(0,2 - 0,15) × 160= 8 (g)$

\(\begin{array}{l}
2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\
c)\\
Theo\,PT:\,n_{H_2SO_4}=n_{H_2}=0,45(mol)\\
\to C\%_{H_2SO_4}=\frac{0,45.98}{230}.100\%≈19,17\%\\
d)\\
n_{Fe_2O_3}=\frac{32}{160}=0,2(mol)\\
Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O\\
Vi\,n_{Fe_2O_3}>\frac{n_{H_2}}{3}\to Fe_2O_3\,du\\
Theo\,PT:\,n_{Fe_2O_3(pứ)}=\frac{1}{3}n_{H_2}=0,15(mol)\\
\to m_{Fe_2O_3(du)}=(0,2-0,15).160=8(g)
\end{array}\)