67)
Phản ứng xảy ra:
\(2Fe + 3C{l_2}\xrightarrow{{{t^o}}}2FeC{l_3}\)
Ta có:
\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol = }}{{\text{n}}_{FeC{l_3}}}\)
\( \to {m_{FeC{l_3}}} = 0,2.(56 + 35,5.3) = 32,5{\text{ gam}}\)
Chọn \(B\).
68)
Phản ứng xảy ra:
\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,3.56 = 16,8{\text{ gam}}\)
\( \to m = {m_{Ag}} = 20 - 16,8 = 3,2{\text{ gam}}\)
Chọn \(A\)
69)
Ta có:
\({n_{S{O_2}}} = \frac{{1,008}}{{22,4}} = 0,045{\text{ mol;}}{{\text{n}}_{NaOH}} = 0,2025.0,4 = 0,081{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,081}}{{0,045}} = 1,8\)
Vì \(1<1,8<2\) nên sản phẩm tạo ra 2 muối là \(Na_2SO_3;NaHSO_3\)
Chọn \(D\).
70)
Phản ứng xảy ra:
\(2Fe + 6{H_2}S{O_4}\xrightarrow{{{t^o}}}F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)
Ta có:
\({n_{Fe}} = \frac{{16,8}}{{56}} = 0,3{\text{ mol}} \to {{\text{n}}_{S{O_2}}} = \frac{3}{2}{n_{Fe}} = 0,45{\text{ mol}}\)
\( \to {V_{S{O_2}}} = 0,45.22,4 = 10,08{\text{ lít}}\)
Chọn \(C\)
