1)
Phản ứng xảy ra:
\(2xR + y{O_2}\xrightarrow{{}}2{R_x}{O_y}\)
BTKL:
\({m_R} + {m_{{O_2}}} = {m_{{R_x}{O_y}}} \to {m_{{O_2}}} = 11,6 - 8,4 = 3,2{\text{ gam}} \to {{\text{n}}_{{O_2}}} = \frac{{3,2}}{{16.2}} = 0,1{\text{ mol}} \to {{\text{n}}_R} = \frac{{2x{n_{{O_2}}}}}{y} = \frac{{0,2x}}{y}\)
\( \to {M_R} = \frac{{8,4}}{{\frac{{0,2x}}{y}}} = 42.\frac{y}{x} \to x = 3;y = 4 \to {M_R} = 56 \to R:F{\text{e}}\)
2)
Z có \(p + e + n = 58;p + n < 40 \to 2p + n = 58;p + n = 40 \to p > 18\)
Ta có \(p \leqslant n \leqslant 1,5p \to 3p \leqslant 2p + n \leqslant 3,5p \to 3p \leqslant 58 \leqslant 3,5p \to 16,57 < p < 19,333 \to p = 19\)
Vậy Z là K (kali).
3)
Ở cốc A xảy ra phản ứng:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({n_{Al}} = \frac{{8,1}}{{27}} = 0,3{\text{ mol;}}{{\text{m}}_{HCl}} = 200.10,95\% = 21,9{\text{ mol}} \to {{\text{n}}_{HCl}} = \frac{{21,9}}{{36,5}} = 0,6{\text{ mol < 3}}{{\text{n}}_{Al}} \to Al\) dư.
\({n_{Al{\text{ phản ứng}}}} = \frac{1}{3}{n_{HCl}} = 0,2{\text{ mol;}}{{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,3{\text{ mol}} \to {{\text{m}}_{dd{\text{ tăng}}}} = 0,2.27 - 0,3.2 = 4,8{\text{ gam}}\)
Ở cốc B xảy ra phản ứng:
\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)
\({m_{{H_2}S{O_4}}} = 200.9,8\% = 19,6{\text{ gam}} \to {n_{{H_2}S{O_4}}} = \frac{{19,6}}{{98}} = 0,2{\text{ mol}}\)
Nếu Zn dư
\( \to {n_{Zn{\text{ tan}}}} = {n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,2{\text{ mol}} \to {{\text{m}}_{dd{\text{ tăng}}}} > 0,2.65 - 0,2.2 = 12,6{\text{ gam}}\)
Vậy axit dư.
Gọi số mol Zn tan là a.
\( \to {n_{Zn}} = {n_{{H_2}}} = a \to 65a - a.2 = 4,8 \to a = \frac{{4,8}}{{63}} \to x = \frac{{4,8}}{{63}}.65 = 4,95{\text{ gam}}\)
