Đáp án:
\(C{\% _{F{e_2}{{(S{O_4})}_3}}} = 12,136\% \)
\(C{\% _{{H_2}S{O_4}}} = 1,784\% \)
Giải thích các bước giải:
Gọi công thức của oxit là \(R_xO_y\)
Ta có:
\( \to {M_{{R_x}{O_y}}} = xR + 16y\)
\( \to \% {m_R} = \frac{{xR}}{{xR + 16y}} = 70\% \to xR = \frac{{112y}}{3} \to R = \frac{{112y}}{{3x}}\)
Thỏa mãn \(x=2;y=3 \to R=56 \to Fe\)
Vậy oxit là \(Fe_2O_3\)
\( \to {n_{F{e_2}{O_3}}} = \frac{{10}}{{56.2 + 16.3}} = 0,0625{\text{ mol}}\)
Cho oxit tác dụng với \(H_2SO_4\)
\(F{e_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}F{e_2}{(S{O_4})_3} + 3{H_2}O\)
\( \to {n_{{H_2}S{O_4}{\text{ phản ứng}}}} = 3{n_{F{e_3}{O_3}}} = 0,1875{\text{ mol}}\)
\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,1875.20\% = 0,0375{\text{ mol}}\)
\( \to {n_{{H_2}S{O_4}{\text{ tham gia}}}} = 0,1875 + 0,0375 = 0,225{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,225.98 = 22,05{\text{ gam}}\)
\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{22,05}}{{11,25\% }} = 196{\text{ gam}}\)
BTKL:
\({m_{dd{\text{ X}}}} = {m_{F{e_3}{O_4}}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} = 10 + 196 = 206{\text{ gam}}\)
\({n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,0625{\text{ mol}}\)
\( \to {m_{F{e_2}{{(S{O_4})}_3}}} = 0,0625.(56.2 + 96.3) = 25{\text{ gam}}\)
\({m_{{H_2}S{O_4}{\text{ dư}}}} = 0,0375.98 = 3,675{\text{ gam}}\)
\( \to C{\% _{F{e_2}{{(S{O_4})}_3}}} = \frac{{25}}{{206}} = 12,136\% \)
\(C{\% _{{H_2}S{O_4}}} = \frac{{3,675}}{{206}} = 1,784\% \)
