Đáp án:
1)
\(m = 12g\)
2)
\({m_{Fe{{\left( {OH} \right)}_2}}} = 13,5g\)
\({m_{NaCl}} = 17,55g\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
{n_{CuS{O_4}}} = V \times {C_M} = 0,1 \times 1 = 0,1mol\\
\Rightarrow {n_{Fe}} = {n_{Cu}} = {n_{CuS{O_4}}} = 0,1mol\\
{m_{Fe\text{ phản ứng}}} = {n_{Fe}} \times M = 0,1 \times 56 = 5,6g\\
{m_{Cu}} = {n_{Cu}} \times M = 0,1 \times 64 = 6,4g\\
m = {n_{Fe}} + {n_{Cu}} - {n_{Fe\text{ phản ứng}}} = 11,2 + 6,4 - 5,6 = 12g\\
2)\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
{n_{FeC{l_2}}} = V \times {C_M} = 0,15 \times 1 = 0,15mol\\
{n_{NaOH}} = V \times {C_M} = 0,3 \times 1,25 = 0,375mol\\
\dfrac{{0,15}}{1} < \dfrac{{0,375}}{2} \Rightarrow \text{dd NaOH dư }\\
\Rightarrow {n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = 0,15mol\\
{n_{NaCl}} = 2 \times {n_{FeC{l_2}}} = 2 \times 0,15 = 0,3mol\\
{m_{Fe{{\left( {OH} \right)}_2}}} = {n_{Fe{{(OH)}_2}}} = 0,15 \times 90 = 13,5g\\
{m_{NaCl}} = {n_{NaCl}} \times M = 0,3 \times 58,5 = 17,55g
\end{array}\)
