a,

$2NaCl+2H_2O\buildrel{{\text{đpmn}}}\over\longrightarrow 2NaOH+H_2+Cl_2$

b,

$n_{NaCl}=\dfrac{58,5}{58,5}=1(kmol)$

Theo PTHH:

$n_{NaOH}=n_{NaCl}=1(kmol)$

$\to m_{NaOH}=1.40.50\%=20kg$

$n_{H_2}=n_{Cl_2}=\dfrac{n_{NaCl}}{2}=0,5(kmol)$

$\to V_{\uparrow}=(0,5+0,5).22,4.50\%=11,2m^3$

Đáp án:

\(\begin{array}{l}
{m_{NaOH}} = 20kg\\
V = 11,2{m^3}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2NaCl + 2{H_2}O \to 2NaOH + C{l_2} + {H_2}\\
b)\\
{m_{NaCl}} = 58,5kg = 58500g\\
{n_{NaCl}} = \dfrac{{58500}}{{58,5}} = 1000\,mol\\
{n_{NaOH}} = {n_{NaCl}} = 1000\,mol\\
{n_{C{l_2}}} = {n_{{H_2}}} = \dfrac{{1000}}{2} = 500\,mol\\
H = 50\% \\
 \Rightarrow {m_{NaOH}} = 1000 \times 50\%  \times 40 = 20000g = 20kg\\
 \Rightarrow V = (500 + 500) \times 50\%  \times 22,4 = 11200l = 11,2{m^3}
\end{array}\)