Đáp số:
a) $9,6g$
b) $C{\% _{N{a_2}S{O_4}}} = 15,62\% ;\\C{\% _{NaCl}} = 5,265\% ;\\C{\% _{NaAl{O_2}}} = 3,28\% $
Giải thích các bước giải:
a) ${n_{FeC{l_3}}} = \dfrac{{19,5}}{{162,5}} = 0,12mol;\\{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{27,36}}{{342}} = 0,08mol$
${n_{{H_2}S{O_4}}} = 0,2mol;{n_{NaOH}} = \dfrac{{77,6}}{{40}} = 1,94mol$
${H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + {H_2}O$
$0,2$ $→$ $0,4$ $0,2$
$FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl$
$0,12$ $→$ $0,36$ $0,12$ $0,36$
$A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}$
$0,08$ $→$ $0,48$ $0,16$ $0,24$
$ \Rightarrow {n_{NaOH(pu)}} = 0,4 + 0,36 + 0,48 = 1,24mol$$ \Rightarrow {n_{NaOH(du)}} = 1,94 - 1,24 = 0,7mol$
Xảy ra phản ứng:
$NaOH + Al{(OH)_3} \to NaAl{O_2} + 2{H_2}O$
$0,16$ $→$ $0,16$ $→$ $0,16$
$ \Rightarrow {n_{NaOH(du)}} = 0,7 - 0,16 = 0,54mol$
Kết tủa $B$ gồm $Fe{(OH)_3}$
$2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O$
$0,12$ $→$ $0,06$
$ \Rightarrow {m_{F{e_2}{O_3}}} = 0,06.160 = 9,6g$
- b)
${m_{dd{H_2}S{O_4}}} = D.V = 200.1,14 = 228g$
Bảo toàn khối lượng:
${m_{FeC{l_3}}} + {m_{A{l_2}{{(S{O_4})}_3}}} + {m_{dd{H_2}S{O_4}}} + {m_{NaOH}} = {m_{ddC}} + {m_{Fe{{(OH)}_3}}}$
$ \Rightarrow {m_{ddC}} = 19,5 + 27,36 + 228 + 77,6 - 0,12.107 = 339,62g$
$ \Rightarrow {m_{{H_2}O}} = 400 - 339,62 = 60,38g$
Dung dịch D gồm:
$N{a_2}S{O_4}(0,44mol);NaCl(0,36mol);NaAl{O_2}(0,16mol)$
$\begin{gathered} \Rightarrow C{\% _{N{a_2}S{O_4}}} = \dfrac{{0,44.142}}{{400}}.100\% = 15,62\% \hfill \\ C{\% _{NaCl}} = \dfrac{{0,36.58,5}}{{400}}.100\% = 5,265\% \hfill \\ C{\% _{NaAl{O_2}}} = \dfrac{{0,16.82}}{{400}}.100\% = 3,28\% \hfill \\ \end{gathered} $
