Đáp án:
${m_{{C_6}{H_5}N{H_3}Cl}} =12,95\,\, gam$
Giải thích các bước giải:
${n_{anilin}} = \frac{{9,3}}{{93}} = 0,1\,\,mol$
Phương trình hóa học:
${C_6}{H_5}N{H_2} + HCl \to {C_6}{H_5}N{H_3}Cl$
$\,\,0,1 \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol$
$\to {m_{{C_6}{H_5}N{H_3}Cl}} = 0,1.129,5 = 12,95\,\, gam$