3)

Phản ứng xảy ra:

\({N_2} + 3{H_2}\xrightarrow{{{t^o},xt}}2N{H_3}\)

 Ta có:

\({n_{N{H_3}}} = \frac{{1,7}}{{14 + 1.3}} = 0,1{\text{ kmol}}\)

\( \to {n_{{N_2}}} = \frac{1}{2}{n_{N{H_3}}} = 0,05{\text{ kmol;}}{{\text{n}}_{{H_2}}} = 3{n_{{N_2}}} = 0,15{\text{ kmol}}\)

\( \to {V_{{N_2}}} = 0,05.22,4 = 1,12{\text{ }}{{\text{m}}^3}\)

\({V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ }}{{\text{m}}^3}\)

4)

Phản ứng xảy ra:

\(4Au + 8NaCN + 2{H_2}O + {O_2}\xrightarrow{{{t^o}}}4NaAu{(CN)_2} + 4NaOH\)

Ta có:

\({m_{Au}} = 500.0,001\%  = 0,005kg = 5{\text{ gam}}\)

\( \to {n_{Au}} = \frac{5}{{197}} \to {n_{NaCn}} = 2{n_{Au}} = \frac{{10}}{{197}}\)

\( \to {m_{NaCN}} = \frac{{10}}{{197}}.(23 + 12 + 14) = 2,4873{\text{ gam}}\)