Đáp án:

Giải thích các bước giải:

3)

\(FeCl_3\): muối: sắt (III) clorua

\(Mg(OH)_2\): bazo: magie hidroxit

\(Ag_2O\): oxit bazo: bạc oxit

\(H_2SO_4\): axit: axit sunfuric

4)

a)

Ta có:

\({n_{CuS{O_4}}} = \frac{{16}}{{64 + 32 + 16.4}} = 0,1{\text{ mol}}\)

\({V_{dd}} = 200{\text{ ml = 0}}{\text{,2 lít}}\)

\( \to {C_{M{\text{ CuS}}{{\text{O}}_4}}} = \frac{{0,1}}{{0,2}} = 0,5M\)

b)

Ta có:

\({m_{{H_2}S{O_4}}} = 150.9,8\%  = 14,7{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{14,7}}{{1.2 + 32 + 16.4}} = 0,15{\text{ mol}}\)

5)

Phản ứng xảy ra:

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

Ta có:

\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol = }}{{\text{n}}_{ZnC{l_2}}}\)

\( \to {m_{ZnC{l_2}}} = 0,2.(65 + 35,5.2) = 27,2{\text{ gam}}\)

\({n_{HCl}} = 2{n_{Zn}} = 0,2.2 = 0,4{\text{ mol}}\)

\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)

\( \to C{\% _{HCl}} = \frac{{{m_{HCl}}}}{{{m_{dd}}}} = \frac{{14,6}}{{182,5}} = 8\% \)