Đáp án:
\(\begin{array}{l}
40)\,C\\
41)\,D\\
42)\,D\\
43)\,D\\
44)\,D\\
45)\,C
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
40)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{{4,2}}{{56}} = 0,075\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,075}}{2} = 0,0375\,mol\\
{m_{F{e_2}{O_3}}} = 0,0375 \times 160 = 6g\\
41)\\
\% N\,trong\,NaN{O_3} = \dfrac{{14}}{{85}} \times 100\% = 16,47\% \\
\% N\,trong\,{(N{H_4})_2}S{O_4} = \dfrac{{28}}{{132}} \times 100\% = 21,21\% \\
\% N\,trong\,N{H_4}N{O_3} = \dfrac{{28}}{{80}} \times 100\% = 35\% \\
\% N\,trong\,{(N{H_2})_2}CO = \dfrac{{28}}{{60}} \times 100\% = 46,67\% \\
44)\\
{M_X} = 3,5 \times 16 = 56dvC \Rightarrow X:Fe\\
\end{array}\)