Đáp án:
\( C{\% _{NaOH}} = 16,077\% \)
Giải thích các bước giải:
Ta có:
\({n_{Na}} = \frac{{23}}{{23}} = 1{\text{ mol}}\)
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({m_{dd{\text{NaOH}}}} = 500.1,2 = 600{\text{ gam}}\)
\( \to {m_{NaOH}} = 600.10\% = 60{\text{ gam}}\)
\( \to {n_{NaOH cũ}} = \frac{{60}}{{40}} = 1,5{\text{ mol}}\)
\({n_{NaOH{\text{ mới}}}} = {n_{Na}} = 1{\text{ mol;}}{{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,5{\text{ mol}}\)
\( \to {n_{NaOH{\text{ trong dd}}{\text{ mới}}}} = 1,5 + 1 = 2,5{\text{ mol}}\)
\( \to {m_{NaOH}} = 2,5.40 = 100{\text{ gam}}\)
BTKL:
\({m_{Na}} + {m_{dd{\text{NaOH}}}} = {m_{dd{\text{mới}}}} + {m_{{H_2}}}\)
\( \to {m_{dd{\text{moi}}}} = 23 + 600 - 0,5.2 = 622{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{{100}}{{622}} = 16,077\% \)
