Đáp án:

\({{\text{m}}_K} = 7,8{\text{ gam;}}{m_{{K_2}O}} = 9,4{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(2K + 2{H_2}O\xrightarrow{{}}2KOH + {H_2}\)

\({K_2}O + {H_2}O\xrightarrow{{}}2KOH\)

Ta có: 

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_K} = 2{n_{{H_2}}} = 0,2{\text{ mol}} \to {{\text{m}}_K} = 0,2.39 = 7,8{\text{ gam}} \to {m_{{K_2}O}} = 17,2 - 7,8 = 9,4{\text{ gam}} \to {{\text{n}}_{{K_2}O}} = \frac{{9,4}}{{39.2 + 16}} = 0,1{\text{ mol}}\)

BTKL: \({m_{hh}} + {m_{{H_2}O}} = {m_{dd{\text{ A}}}} + {m_{{H_2}}} \to 17,2 + 84,8 = {m_{dd{\text{ A}}}} + 0,1.2 \to {m_{dd{\text{ A}}}} = 101,8{\text{ gam}}\)

\({n_{KOH}} = {n_K} + 2{n_{{K_2}O}} = 0,2 + 0,1.2 = 0,4{\text{ mol}} \to {{\text{m}}_{KOH}} = 0,4.56 = 22,4{\text{ gam}} \to {\text{C}}{{\text{\% }}_{KOH}} = \frac{{22,4}}{{101,8}} = 22\% \)

n H2=$\frac{2,24}{22,4}$=0,1 mol

2K+2H2O→2KOH+H2↑

0,2                      ←0,1          mol

K2O+H2O→2KOH

0,1→             0,2             mol

a.

m K=0,2.39=7,8 g

m K2O=17,2-7,8=9,4 g

b.

n K2O=$\frac{9,4}{94}$=0,1 mol

∑n KOH=0,2+0,2=0,4 mol

m ct KOH=0,4.56=22,4 g

mdd sau=m hh+m H2O-m H2

              =17,2+84,8-0,1.2=101,8 g

⇒C% KOH=$\frac{22,4}{101,8}$.100≈22 %

-----------------Nguyễn Hoạt----------------