Đáp án:
${V_{{N_2}}} = 44,8\,\,l $
$ {V_{{H_2}}} = 134,4\,\,\,l $
$V =332\,\,ml $
Giải thích các bước giải:
${n_{N{H_3}}} = \dfrac{{34}}{{18}} = 2\,\,mol $
Phương trình hóa học:
$ {N_2} + 3{H_2} \to 2N{H_3} $
$ 1\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\, \leftarrow 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
Ta có: $ H = 50\% $
$ \to {n_{{N_2}}} = 1:50\% = 2\,\,mol \to {V_{{N_2}}} = 44,8\,\,l $
$ \,\,\,\,\,\,{n_{{H_2}}} = 3:50\% = 6\,\,mol \to {V_{{H_2}}} = 134,4\,\,\,l $
Phương trình hóa học:
$ N{H_3} + HCl \to N{H_4}Cl $
$ 2\,\,\,\,\,\,\,\, \to \,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$ \to {n_{HCl}} = 2\,\,mol $
$ \to {m_{HCl}} = 2.36,5 = 73\,\,gam $
$ \to {m_{d\,d\,\,HCl}} = 73:20\% = 365\,\,g $
$ \to V = \dfrac{m}{d} = \dfrac{{365}}{{1,1}} = 332\,\,ml $
