Đáp án:
$PTHH:2X+2nH_2O\to 2X(OH)_n+nH_2$
$n_{H_2}=\dfrac{0.448}{22.4}=0.02(mol)$
Theo pt: $n_X=\dfrac{2}{n}n_{H_2}=\dfrac{0.04}{n}(mol)$
$M_X=\dfrac{1.56}{\dfrac{0.04}{n}}=39n$
Xét $n=1\to M_X=39(K)$
Xét $n=2\to M_X=78(\text{Loại})$
$n_K=\dfrac{1.56}{39}=0.04(mol)$
$PTHH:2K+2H_2O\to 2KOH+H_2$
Theo pt: $n_{KOH}=n_K=0.04(mol)$
$C_{M_{KOH}}=\dfrac{0.04}{0.2}=0.2(M)$
$n_{H_2SO_4}=\dfrac{20×19.6}{100×98}=0.04(mol)$
Lập tỉ lệ: $\dfrac{n_{KOH}}{n_{H_2SO_4}}=\dfrac{0.04}{0.04}=1$
$\to$ Phản ứng tạo $KHSO_4$
$PTHH:KOH+H_2SO_4\to KHSO_4+H_2O$
Theo pt: $n_{KHSO_4}=n_{KOH}=0.4(mol)$