cho 10,8 gam hon hop fe va cuo vao h2so4 dac nong thu duoc 1,68 lit dkc
A/ viet phuong trinh phan ung xay ra
B/ tinh phan tram khoi luong fe, cuo
C/ tinh khoi luong muoi tao thanh
cho 10,8 gam hon hop fe va cuo vao h2so4 dac nong thu duoc 1,68 lit dkc
A/ viet phuong trinh phan ung xay ra
B/ tinh phan tram khoi luong fe, cuo
C/ tinh khoi luong muoi tao thanh
Đáp án:
b) \(\% {m_{Fe}} = 25,9\% ;\% {m_{CuO}} = 74,1\% \)
c) \({{\text{m}}_{hh{\text{ muối}}}} = 26{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Fe + 6{H_2}S{O_4}\xrightarrow{{}}F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)
\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)
Ta có:
\({n_{S{O_2}}} = \frac{{1,68}}{{22,4}} = 0,075{\text{ mol}} \to {{\text{n}}_{Fe}} = \frac{2}{3}{n_{S{O_2}}} = 0,05{\text{ mol}}\)
\({m_{Fe}} = 0,05.56 = 2,8{\text{ gam}} \to {{\text{m}}_{CuO}} = 8{\text{ gam}} \to {{\text{n}}_{CuO}} = \frac{8}{{64 + 16}} = 0,1{\text{ mol}}\)
\(\% {m_{Fe}} = \frac{{2,8}}{{10,8}} = 25,9\% \to \% {m_{CuO}} = 74,1\% \)
\({n_{F{e_2}{{(S{O_4})}_3}}} = \frac{1}{2}{n_{Fe}} = 0,025{\text{ mol; }}{{\text{n}}_{CuS{O_4}}} = {n_{CuO}} = 0,1{\text{ mol}} \to {{\text{m}}_{hh{\text{ muối}}}} = 0,025.(56.2 + 96.3) + 0,1.160 = 26{\text{ gam}}\)