Đáp án:
Bạn tham khảo lời giải ở dưới nhé !
Giải thích các bước giải:
Câu 3:
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
3C{l_2} + 6KOH \to 5KCl + KCl{O_3} + 3{H_2}O
\end{array}\)
\(\begin{array}{l}
a)\\
{n_{KMn{O_4}}} = 0,02mol\\
\to {n_{C{l_2}}} = \dfrac{5}{2}{n_{KMn{O_4}}} = 0,05mol\\
\to {V_{C{l_2}}} = 1,12l\\
{n_{KOH}} = 0,8mol\\
\to \dfrac{{{n_{C{l_2}}}}}{3} < \dfrac{{{n_{KOH}}}}{6} \to {n_{KOH}}d{\rm{u}}\\
\to {n_{KOH}} = 2{n_{C{l_2}}} = 0,1mol\\
\to {n_{KOH(dư)}} = 0,7mol
\end{array}\)
\(\begin{array}{l}
{n_{KCl}} = \dfrac{5}{3}{n_{C{l_2}}} = 0,083mol\\
{n_{KCl{O_3}}} = \dfrac{1}{3}{n_{C{l_2}}} = 0,017mol\\
\to C{M_{KOH(du)}} = \dfrac{{0,7}}{{0,8 + 1,12}} = 0,36M\\
\to C{M_{KCl}} = \dfrac{{0,083}}{{0,8 + 1,12}} = 0,04M\\
\to C{M_{KCl{O_3}}} = \dfrac{{0,017}}{{0,8 + 1,12}} = 0,009M
\end{array}\)
\(\begin{array}{l}
b)\\
KOH + HCl \to KCl + {H_2}O\\
KCl{O_3} + 6HCl \to KCl + 3C{l_2} + 3{H_2}O
\end{array}\)
\(\begin{array}{l}
{n_{HCl}} = {n_{KOH(dư)}} + 6{n_{KCl{O_3}}} = 0,802mol\\
\to {m_{HCl}} = 29,27g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{29,27}}{{35\% }} \times 100\% = 83,63g
\end{array}\)
Câu 4:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{m_{HCl}} = \dfrac{{200 \times 18,25\% }}{{100\% }} = 36,5g\\
\to {n_{HCl}} = 1mol
\end{array}\)
Gọi a và b lần lượt là số mol của Al và Fe
\(\begin{array}{l}
\left\{ \begin{array}{l}
27a + 56b = 11\\
3a + 2b = 1
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,3\\
b = 0,05
\end{array} \right.\\
\to {n_{Al}} = 0,3mol\\
\to {n_{Fe}} = 0,05mol
\end{array}\)
\(\begin{array}{l}
a)\\
\% {m_{Al}} = \dfrac{{0,3 \times 27}}{{11}} \times 100\% = 73,64\% \\
\% {m_{Fe}} = 100\% - 73,64\% = 26,36\%
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,3mol\\
\to {m_{AlC{l_3}}} = 40,05g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,05mol\\
\to {m_{FeC{l_2}}} = 6,35g\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} + {n_{Fe}} = 0,5mol\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 210g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{40,05}}{{210}} \times 100\% = 19,07\% \\
\to C{\% _{FeC{l_2}}} = \dfrac{{6,35}}{{210}} \times 100\% = 3,02\%
\end{array}\)
c,
Dung dịch X gồm \(AlC{l_3},FeC{l_2}\)
\(\begin{array}{l}
2FeC{l_2} + C{l_2} \to 2FeC{l_3}\\
{n_{C{l_2}}} = 0,025mol\\
{n_{C{l_2}}} = \dfrac{1}{2}{n_{FeC{l_2}}} = 0,025mol
\end{array}\)
Suy ra khí \(C{l_2}\) phản ứng vừa đủ
Suy ra dung dịch Y gồm: \(AlC{l_3},FeC{l_3}\)
Để kết tủa lớn nhất khi \(Al{(OH)_3}\) không tan trong NaOH dư
\(\begin{array}{l}
{n_{FeC{l_3}}} = {n_{FeC{l_2}}} = 0,05mol\\
AlC{l_3} + 3NaOH \to Al{(OH)_3} + 3NaCl\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
{n_{NaOH}} = 3{n_{AlC{l_3}}} + 3{n_{FeC{l_3}}}\\
\to {n_{NaOH}} = 1,05mol\\
\to {V_{NaOH}} = 1,05M\\
a = {m_{Al{{(OH)}_3}}} + {m_{Fe{{(OH)}_3}}}\\
\to a = 0,3 \times 78 + 0,05 \times 107 = 28,75g
\end{array}\)
