Đáp án:

$2K + 2H_{2}O→2KOH + H_{2}↑$ (1)

$K_{2}O + H_{2}O→2KOH$ (2)

a) Ta có: $n_{H_{2}}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}= 0,3 (mol)$

Theo PT (1): $n_{K}=2n_{H_{2}}=2 × 0,3= 0,6 (mol)$

⇒ $m_{K}=n × M=0,6 × 39= 23,4 (g)$

$m_{hh}=m_{K} + m_{K_{2}O}$

⇒ $m_{K_{2}O}=m_{hh} - m_{K}=32,8 - 23,4= 9,4 (g)$

$\text{%$m_{K}=\dfrac{23,4}{32,8}$ × 100%= 71,34%}$

$\text{%$m_{K_{2}O}$=100 - 71,34= 28,66%}$

b) $n_{K_{2}O}=\dfrac{m}{M}=\dfrac{9,4}{94}= 0,1 (mol)$

Theo PT (1): $n_{KOH}=2n_{H_{2}}=2 × 0,3= 0,6 (mol)$

Theo PT (2): $n_{KOH}=2n_{K_{2}O}=2 × 0,1= 0,2 (mol)$

⇒ ∑$n_{KOH}=0,6 + 0,2= 0,8 (mol)$

⇒ $m_{KOH}=n × M=0,8 × 56= 44,8 (g)$

 

`#Ly`

$a)$

$n_{H_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{6,72}{22,4}$ $=$ $0,3$ $(mol)$

Do chỉ có $K$ tác dụng với nước tạo khí $H_2$, nên ta có:

PTHH: $2K$ $+$ $2H_2O$ `->` $2KOH$ $+$ $H_2↑$

Theo PT, $n_{K}$ $=$ $n_{KOH}$ $=$ $2n_{H_2}$ $=$ $0,6$ $(mol)$

$m_{K}$ $=$ $n.M=0,6.39=23,4$ $(g)$

`%_(K` $=$ $\dfrac{m_K}{m_{hh}}$`.100%` $=$ $\dfrac{23,4}{32,8}$`.100%` $≈$ $71,34$`%`

`%_(K_2O` $=$ $100-71,34=28,66$`%`

$b)$

PTHH: $K_2O$ $+$ $H_2O$ `->` $2KOH$  

$m_{K_2O}$ $=$ $32,8-23,4=9,4$ $(g)$

$n_{K_2O}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{9,4}{94}$ $=$ $0,1$ $(mol)$

Theo PT trên, $n_{KOH}$ $=$ $2n_{K_2O}$ $=$ $0,2$ $(mol)$

$\sum\limits_{}^{}$$m_{KOH}$ $=$ $56.(0,6+0,2)=44,8$ $(g)$