Đáp án + Giải thích các bước giải:

$a\big)2K+2H_2O\to 2KOH+H_2$

$b\big)$

$n_K=\frac{3,9}{39}=0,1(mol)$

Theo PT: $n_{H_2O}=n_K=0,1(mol)$

$\to m_{H_2O}=0,1.18=1,8(g)$

$c\big)$

Theo PT: $n_{KOH}=n_K=0,1(mol)$

$\to m_{KOH}=0,1.56=5,6(g)$

$d\big)$

Theo PT: $n_{H_2}=\frac{1}{2}n_K=0,05(mol)$

$\to m_{dd\,spu}=3,9+50-0,05.1=53,8(g)$

$\to C\%_{KOH}=\frac{5,6}{53,8}.100\%\approx 10,41\%$

Đáp án + Giải thích các bước giải:

 a) PTHH:
$2K + 2H_2O \longrightarrow 2KOH + H_2 \uparrow$
b) $n_K = \dfrac{m}{M}= \dfrac{3,9}{39} = 0,1\text{ mol}$

Theo PTHH:

$n_{H_2O} =n_K = 0,1\text{ mol}$
$\Rightarrow m_{H_2O} = n . M = 0,1 . (2 . 1 + 16) = 0,1 . 18 = 1,8g$

c) Theo PTHH:
$n_{KOH} = n_K = 0,1\text{ mol}$
$\Rightarrow m_{KOH} = n . M = 0,1 . (39 + 16 + 1) = 0,1 . 56 = 5,6g$

d) Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_K = \dfrac{1}{2} . 0,1 = 0,05\text{ mol}$

$\Rightarrow m_{H_2}=  n . M = 0,05 . 2 = 0,1g$

$\Rightarrow m_{ddKOH} = m_K + m_{H_2O} - m_{H_2} = 3,9 + 50 - 0,1 = 53,8g$

$\Rightarrow \%_{KOH}=  \dfrac{m_{ct}}{m_{dd}} .100\%= \dfrac{5,6}{53,8} . 100\% \approx 10,41\%$