Đáp án:
a) Fe: 36,84%; Cu: 63,16%
b) 126 g
c) 255 g
Giải thích các bước giải:
a) ${n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol$
Gọi x, y là số mol $Fe$, $Cu$
Ta có hpt: $\left\{ \begin{gathered}
56x + 64y = 15,2 \hfill \\
3x + 2y = 0,3.2(BTe) \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
x = 0,1 \hfill \\
y = 0,15 \hfill \\
\end{gathered} \right.$
$\begin{gathered}
\Rightarrow \% {m_{Fe}} = \dfrac{{0,1.56}}{{15,2}}.100\% = 36,84\% \hfill \\
\% {m_{Cu}} = 100 - 36,84 = 63,16\% \hfill \\
\end{gathered} $
b) Bảo toàn nguyên tố $S$:
$\begin{gathered}
{n_{{H_2}S{O_4}}} = {n_{CuS{O_4}}} + {n_{S{O_2}}} + 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,15 + 0,3 + 0,1.3 = 0,75mol \hfill \\
\Rightarrow {n_{{H_2}S{O_4}(tt)}} = \dfrac{{0,75.105}}{{100}} = 0,7875mol \hfill \\
\Rightarrow {V_{dd{H_2}S{O_4}}} = \dfrac{{0,7875}}{8} = 0,0984\left( l \right) = 98.4ml \hfill \\
\Rightarrow {m_{dd{H_2}S{O_4}}} = V.D = 98,4.1,28 = 126g \hfill \\
\end{gathered} $
c)
Gọi oleum là ${H_2}S{O_4}.nS{O_3}$
$\begin{gathered}
{H_2}S{O_4}.nS{O_3} + n{H_2}O \to (n + 1){H_2}S{O_4} \hfill \\
{H_2}S{O_4} + N{a_2}C{O_3} \to N{a_2}S{O_4} + C{O_2} + {H_2}O \hfill \\
\end{gathered} $
⇒ ${n_{oleum}} = {n_{C{O_2}}} = 0,5mol $
$\begin{gathered}
{n_{oleum}} = \dfrac{{0,5}}{{n + 1}} \Rightarrow {M_{oleum}} = 85\left( {n + 1} \right) \hfill \\
\Rightarrow 98 + 80n = 85n + 85 \Rightarrow n = 2,6 \hfill \\
\end{gathered} $
A là ${H_2}S{O_4}.2,6S{O_3}$
${H_2}S{O_4}.2,6S{O_3} + 2,6{H_2}O \to 3,6{H_2}S{O_4}$
Gọi x là số mol của oleum A
${m_{dd{H_2}S{O_4}}} = 500.1,28 = 640g \Rightarrow {m_{{H_2}S{O_4}}} = 64g$
$\begin{gathered}
C{\% _{{H_2}S{O_4}sau}} = \dfrac{{3,6x.98 + 64}}{{306x + 640}}.100\% = 40\% \hfill \\
\Rightarrow x = \dfrac{5}{6} \hfill \\
\end{gathered} $
${m_A} = 306.\dfrac{5}{6} = 255g$