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@phamnghi9674

#NOCOPY

7)

\({n_{NaOH}} = 0,25.0,5 = 0,125{\text{ mol}}\)

\( \to {V_{dd{\text{ NaOH 2M}}}} = \frac{{0,125}}{2} = 0,0625{\text{ lit = 62}}{\text{,5 ml}}\)

\( \to {m_{{H_2}O}} = 250 - 62,5 = 187,5{\text{ ml}}\)

8)

Phản ứng xảy ra:

\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)

Ta có:

\({m_{NaOH}} = 200.20\%  = 40{\text{ gam}}\)

\( \to {n_{NaOH}} = {n_{HCl}} = {n_{NaCl}} = \frac{{40}}{{23 + 16 + 1}} = 1{\text{ mol}}\)

BTKL:

\({m_{dd}} = {m_{dd\;NaOH}} + {m_{{\text{dd   }}HCl}} = 200 + 100 = 300{\text{ gam}}\)

\({m_{NaCl}} = 1.(23 + 35,5) = 58,5{\text{ gam}}\)

\( \to C{\% _{NaCl}} = \frac{{58,5}}{{300}} = 19,5\% \)

\({m_{HCl}} = 1.(1 + 35,5) = 36,5{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{36,5}}{{100}} = 36,5\% \)