Đáp án:
$a,$ $V_{H_2}$ = $1,68(l)$

$b,$ $\text{C% $H_2SO_4$ =  3,675(%)}$

$nAl$ = $\frac{1,35}{27}$ = $0,05(mol)$

#Nếu là $H_2SO_4$ loãng thì có phản ứng:

$2Al$ + $3H_2SO_4$ → $Al_2(SO_4)_3$ + $3H_2$

$a,$

$→$ $nH_2$ = $\frac{3}{2}$ $nAl$ = $\frac{3}{2}$ * $0,05$ = $0,075(mol)$

$→$ $V_{H_2}$ = $0,075 * 22,4$ = $1,68(l)$

$b,$

$nH_2SO_4$ = $nH_2$ = $0,075(mol)$

$\text{C% $H_2SO_4$ = $\frac{0,075 * 98}{200}$ * 100 = 3,675(%)}$

Đáp án + Giải thích các bước giải:

Câu 6:

$a\big)$

$n_{Al}=\frac{1,35}{27}=0,05(mol)$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

Theo PT: $n_{H_2}=\frac{3}{2}n_{Al}=0,075(mol)$

$\to V_{H_2(đktc)}=0,075.22,4=1,68(l)$

$b\big)$

Theo PT: $n_{H_2SO_4}=n_{H_2}=0,075(mol)$

$\to C\%_{H_2SO_4}=\frac{0,075.98}{200}.100\%=3,675\%$