Đáp án:
a) 5,7g
b) 4,294g
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{HCl}} = 0,38 \times 0,2 = 0,076mol\\
{n_{{H_2}O}} = \dfrac{{{n_{HCl}}}}{2} = 0,038\,mol\\
BTKL:\\
{m_A} + {m_{HCl}} = m + {m_{{H_2}O}} \Leftrightarrow m = {m_A} + {m_{HCl}} - {m_{{H_2}O}}\\
\Leftrightarrow m = 3,61 + 0,076 \times 36,5 - 0,038 \times 18 = 5,7g\\
b)\\
FeC{l_3} + 3NaOH \to Fe{(OH)_3} + 3NaCl\\
ZnC{l_2} + 2NaOH \to Zn{(OH)_2} + 2NaCl\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
{n_{NaOH}} = {n_{HCl}} = 0,076\,mol\\
{n_{NaCl}} = {n_{NaOH}} = 0,076\,mol\\
BTKL:\\
{m_{hh}} + {m_{NaOH}} = {m_ \downarrow } + {m_{NaCl}} \Leftrightarrow {m_ \downarrow } = {m_{hh}} + {m_{NaOH}} - {m_{NaCl}}\\
\Leftrightarrow {m_ \downarrow } = 5,7 + 0,076 \times 40 - 0,076 \times 58,5 = 4,294g
\end{array}\)
