Đáp án:

$30,03\%$

Giải thích các bước giải:

\(\begin{array}{l}
2Na + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{m_{{H_2}S{O_4}}} = \dfrac{{a \times C}}{{100}} = 0,01aC\,(g)\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,01aC}}{{98}} = \dfrac{{aC}}{{9800}}(mol)\\
{m_{{H_2}O}} = a - 0,01aC(g)\\
{n_{{H_2}O}} = \dfrac{{a - 0,01aC}}{{18}}\,mol\\
{n_{{H_2}}} = \dfrac{{aC}}{{9800}} + \dfrac{{a - 0,01aC}}{{18}} \times \dfrac{1}{2} = \dfrac{{aC}}{{9800}} + \dfrac{{a - 0,01aC}}{{36}}(mol)\\
 \Rightarrow 2 \times (\dfrac{{aC}}{{9800}} + \dfrac{{a - 0,01aC}}{{36}}) = 0,045a\\
 \Rightarrow \dfrac{{aC}}{{4900}} + \dfrac{{a - 0,01aC}}{{18}} = 0,045a\\
 \Rightarrow 9aC + 2450(a - 0,01aC) = 1984,5a\\
 \Rightarrow 9aC - 24,5aC + 2450a = 1984,5a\\
 \Rightarrow 465,5a = 15,5aC \Rightarrow C = 30,03
\end{array}\)