Đáp án:
\({m_{Fe}} = 16,8{\text{ gam}}\)
\({m_{Cu}} = 6,4{\text{ gam}}\)
\({V_{S{O_2}}} = 12,32{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{Fe}}\)
\( \to {m_{Fe}} = 0,3.56 = 16,8{\text{ gam}}\)
\( \to {m_{Cu}} = 23,2 - 16,8 = 6,4{\text{ gam}}\)
\({n_{Cu}} = \frac{{6,4}}{{64}} = 0,1{\text{ mol}}\)
Bảo toàn e:
\(3{n_{Fe}} + 2{n_{Cu}} = 2{n_{S{O_2}}}\)
\( \to 0,3.3 + 0,1.2 = 2{n_{S{O_2}}}\)
\( \to {n_{S{O_2}}} = 0,55{\text{ mol}}\)
\( \to {V_{S{O_2}}} = 0,55.22,4 = 12,32{\text{ lít}}\)