Đáp án+ Giải thích các bước giải:

$a)$

$-)PTHH:$ $Ba + 2H_2O$ $\rightarrow$ $Ba(OH)_2 + H_2(1)$

$2K + 2H_2O$ $\rightarrow$ $2KOH + H_2(2)$ $b)$

Gọi $a, b$ lần lượt là $n_{Ba}$ , $n_{K}$ $(a,b>0)$

Ta có:  $\begin{cases} 137a+39b=33,25\\a+0,5b= \dfrac{6,16}{22,4}\end{cases}$ $\rightarrow$ $\begin{cases} a=0,2\\b=0,1 \end{cases}$

$\rightarrow$ $\text{% $m_{Ba}$= $\frac{0,2.137.100}{33,25}$= 82,4% }$

$\rightarrow$ $\text{% $m_{K}$= 100-82,4=17,6% }$

$c)$

Theo pt $(1)$ : $n_{Ba(OH)_2}$= $n_{Ba}$= $0,2(mol)$

Theo pt $(2)$ : $n_{KOH}$= $n_{K}$= $0,15(mol)$

$m_{bazơ}$= $m_{Ba(OH)_2}$ + $m_{KOH}$ = $0,2.171+0,15.56=42,6(g)$

$\rightarrow$ $\text{% $m_{Ba(OH)_2}$= $\frac{0,2.171.100}{42,6}$= 80,28% }$

$\rightarrow$ $\text{% $m_{KOH}$= 100-80,28=19,72% }$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{6,16}}{{22,4}} = 0,275\,mol\\
hh:Ba(a\,mol),K(b\,mol)\\
\left\{ \begin{array}{l}
137a + 39b = 33,25\\
a + 0,5b = 0,275
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,15\\
\% {m_{Ba}} = \dfrac{{0,2 \times 137}}{{33,25}} \times 100\%  = 82,41\% \\
\% {m_K} = 100 - 82,41 = 17,59\% \\
c)\\
{n_{Ba{{(OH)}_2}}} = {n_{Ba}} = 0,2\,mol\\
{n_{KOH}} = {n_K} = 0,15\,mol\\
\% {m_{Ba{{(OH)}_2}}} = \dfrac{{0,2 \times 171}}{{0,2 \times 171 + 0,15 \times 56}} \times 100\%  = 80,28\% \\
\% {m_{KOH}} = 100 - 80,28 = 19,72\% 
\end{array}\)