Câu 3:

1) 

Ta có:

\({n_{NaOH}} = 0,2.0,5 = 0,1{\text{mol}}\)

2)

Ta có:

\({m_{KOH}} = 300.11,2\%  = 33,6{\text{ gam}} \to {{\text{n}}_{KOH}} = \frac{{33,6}}{{56}} = 0,6{\text{ mol}}\)

3)

Ta có:

\({m_{{H_2}S{O_4}}} = 200.19,6 = 39,2{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{39,2}}{{98}} = 0,4{\text{ mol}}\)

4)

\({V_{dd{\text{ NaOH}}}} = \frac{{240}}{{1,2}} = 200{\text{ ml = 0,2 lít}}\)

\( \to {n_{NaOH}} = 0,2.0,2 = 0,04{\text{ mol}}\)

Câu 4)

1) 

Ta có:

\({m_{dd}} = 5 + 120 = 125{\text{ gam}} \to {\text{C}}{{\text{\% }}_{NaCl}} = \frac{5}{{125}}.100\%  = 4\% \)

2)

Ta có:

\({n_{NaOH}} = \frac{{16}}{{40}} = 0,4{\text{ mol}} \to {{\text{C}}_{M{\text{ NaOH}}}} = \frac{{0,4}}{{0,2}} = 2M\)

3)

\({n_{CuS{O_4}}} = \frac{{4,8}}{{64 + 96}} = 0,03{\text{ mol}} \to {{\text{C}}_{M{\text{ CuS}}{{\text{O}}_4}}} = \frac{{0,03}}{{0,2}} = 0,15M\)

5)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{32,5}}{{65}} = 0,5{\text{ mol = }}{{\text{n}}_{{H_2}}} = {n_{ZnC{l_2}}}\)

\( \to {V_{{H_2}}} = 0,5.22,4 = 11,2{\text{ lít}}\)

\({m_{ZnC{l_2}}} = 0,5.(65 + 35,5.2) = 68{\text{ gam}}\)