Đáp án:
a) \(\to {{\text{m}}_{K{\text{ tham gia}}}} = 12,1875{\text{ gam}}\)
b) \({{\text{V}}_{C{l_2}{\text{ tham gia}}}} = 3,5{\text{ lít}}\)
Giải thích các bước giải:
\(2K + C{l_2}\xrightarrow{{}}2KCl\)
Ta có:
\({n_{KCl}} = \frac{{18,625}}{{39 + 35,5}} = 0,25{\text{ mol = }}{{\text{n}}_{K{\text{ phản ứng}}}} \to {n_{K{\text{ tham gia}}}} = \frac{{0,25}}{{80\% }} = 0,3125{\text{ mol}} \to {{\text{m}}_{K{\text{ tham gia}}}} = 0,3125.39 = 12,1875{\text{ gam}}\)
\({n_{C{l_2}{\text{ phản ứng}}}} = \frac{1}{2}{n_{KCl}} = 0,125{\text{ mol}} \to {{\text{n}}_{C{l_2}{\text{ tham gia}}}} = \frac{{0,125}}{{80\% }} = 0,15625{\text{ mol}} \to {{\text{V}}_{C{l_2}{\text{ tham gia}}}} = 0,15625.22,4 = 3,5{\text{ lít}}\)
