Đáp án:
X là Zn (kẽm)
Giải thích các bước giải:
Gọi n là hóa trị của X.
\(4X + n{O_2}\xrightarrow{{{t^o}}}{X_2}{O_n}\)
\(2X + nC{l_2}\xrightarrow{{}}2XC{l_n}\)
BTKL:
\({m_{{\text{C}}{{\text{l}}_2}{\text{}}}} = 5,97 - 4,55 = 1,42{\text{ gam}} \to {{\text{n}}_{C{l_2}}} = \frac{{1,42}}{{35,5.2}} = 0,02{\text{ mol}}\)
\({m_{{O_2}}} = 6,77 - 5,97 = 0,8{\text{ gam}} \to {{\text{n}}_{{O_2}}} = 0,025{\text{ mol}}\)
Ta có:
\({n_X} = \frac{{4{n_{{O_2}}}}}{n} + \frac{{2{n_{C{l_2}}}}}{n} = \frac{{0,025.4}}{n} + \frac{{0,02.2}}{n} = \frac{{0,14}}{n} \to {M_X} = \frac{{4,55}}{{\frac{{0,14}}{n}}} = 32,5n \to n = 2;{M_X} = 65 \to X:Zn\)
Ta có:
\({n_{ZnC{l_2}}} = {n_{C{l_2}}} = 0,02{\text{ mol}} \to {{\text{m}}_{ZnC{l_2}}} = 0,02.(65 + 35,5.2) = 2,72{\text{ gam}} \to {\text{\% }}{{\text{m}}_{ZnC{l_2}}} = \frac{{2,72}}{{6,77}} = 40,18\% \to \% {m_{ZnO}} = 59,82\% \)
