6)
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{9,2}}{{23}} = 0,4{\text{ mol = }}{{\text{n}}_{NaOH}}\)
\({n_{{H_2}}} = \frac{1}{2}{n_{Na}} = \frac{1}{2}.0,4 = 0,2{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({m_{NaOH}} = 0,4.(23 + 16 + 1) = 16{\text{ gam}}\)
7)
a)
\({n_{CuS{O_4}}} = \frac{{400}}{{64 + 32 + 16.4}} = 2,5{\text{ mol}}\)
\( \to {C_{M{\text{ CuS}}{{\text{O}}_4}}} = \frac{{2,5}}{4} = 0,625M\)
b)
\({C_{M{\text{ MgC}}{{\text{l}}_2}}} = \frac{{0,5}}{{1,5}} = 0,333M\)
c)
\({V_{dd}} = 1500{\text{ ml = 1}}{\text{,5 lít}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}C{O_3}}} = \frac{{0,06}}{{1,5}} = 0,04M\)
8)
a)
\(C{\% _{KCl}} = \frac{{{m_{KCl}}}}{{{m_{dd}}}} = \frac{{20}}{{600}} = 3,33\% \)
b)
\(C{\% _{{K_2}S{O_4}}} = \frac{{{m_{{K_2}S{O_4}}}}}{{{m_{dd}}}} = \frac{{75}}{{1500}} = 5\% \)
c)
\({m_{dd}} = {m_{NaCl}} + {m_{{H_2}O}} = 15 + 45 = 60{\text{ gam}}\)
\( \to C{\% _{NaCl}} = \frac{{15}}{{60}} = 25\% \)
d)
\({n_{HCl}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol}}\)
\( \to {m_{HCl}} = 0,2.36,5 = 7,3{\text{ gam}}\)
\({m_{dd}} = 7,3 + 500 = 507,3{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{7,3}}{{507,3}} = 1,43\% \)