Đáp án:
a) 466 g
b) 78,4%
c) 300 g
Giải thích các bước giải:
a) ${n_{BaC{l_2}}} = \dfrac{{4160.10\% }}{{208}} = 2mol$
${H_2}S{O_4}$ dư ⇒ ${n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 2mol \Rightarrow {m_{BaS{O_4}}} = 466g$
b)
${n_{{H_2}S{O_4}(pu)}} = {n_{BaS{O_4}}} = 2mol;{n_{{H_2}S{O_4}(du)}} = \dfrac{1}{2}{n_{KOH}} = 2mol$
$\begin{gathered}
\Rightarrow \sum {{n_{{H_2}S{O_4}}} = 2 + 2 = 4mol} \hfill \\
\Rightarrow C{\% _{{H_2}S{O_4}}} = \dfrac{{4.98}}{{500}}.100\% = 78,4\% \hfill \\
\end{gathered} $
c) $Fe{S_2} \to 2{H_2}S{O_4}$
${n_{Fe{S_2}(pu)}} = \dfrac{1}{2}{n_{{H_2}S{O_4}}} = 2mol$
$H = 80\% \Rightarrow {n_{Fe{S_2}(tt)}} = \dfrac{{2.100}}{{80}} = 2,5mol$
$ \Rightarrow {m_{Fe{S_2}}} = 2,5.120 = 300g$