Đáp án:
PTHH: \[2M+3H_2SO_4\to M_2{(SO_4)}_3+3H_2\]
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
Theo PTHH: $n_M=\dfrac{2}{3}. n_{H_2}=0,2(mol)$
$\to M_M=\dfrac{5,4}{0,2}=27(g/mol)$
$\to$ M là Nhôm
$n_{H_2SO_4}=n_{H_2}=0,3(mol)$
$\to m=\dfrac{0,3.98}{9,8\%}=300(g)$