Đáp án + Giải thích các bước giải:

\(\begin{array} {l} n_{K_2O}=\dfrac{47}{94}=0,5(mol)\\ K_2O+H_2O\to 2KOH\\ n_{KOH(pứ)}=2n_{K_2O}=1(mol)\\ m_{KOH(pứ)}=1.56=56(g)\\ \text{Đặt }m_{dd\,KOH\,bđ}=x(g)\\ \to C\%_{KOH\,sau}=\dfrac{x.7,93\%+56}{x+56}.100\%=21\%\\ \to x\approx 352,9(g)\\ \to m_{dd\,KOH\,7,93\%}\approx 352,9(g) \end{array}\)

Em tham khảo nha :

\(\begin{array}{l}
{K_2}O + {H_2}O \to 2KOH\\
{n_{{K_2}O}} = \dfrac{{47}}{{94}} = 0,5mol\\
{n_{KOH}} = 2{n_{{K_2}O}} = 1mol\\
{m_{KOH}} = 1 \times 56 = 56g\\
KOH\,7,93\% :\\
{m_{{\rm{dd}}KOH}} = a(g)\\
{m_{KOH}} = 0,0793a(g)\\
C{\% _{KOH}} = \dfrac{{0,0793a + 56}}{{a + 47}} \times 100\%  = 21\% \\
 \Rightarrow \dfrac{{0,0793a + 56}}{{a + 47}} = 0,21\\
 \Rightarrow a = 353g
\end{array}\)