Giải thích các bước giải:
Câu 4:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{HCl}} = 0,2mol\\
\to {n_{Zn}} = {n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
\to {m_{Zn}} = 0,1 \times 65 = 6,5g\\
b)\\
{n_{ZnC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,1mol\\
\to {m_{ZnC{l_2}}} = 0,1 \times 136 = 13,6g\\
{m_{{\rm{dd}}HCl}} = 200 \times 1,1 = 220g\\
\to {m_{{\rm{dd}}ZnC{l_2}}} = {m_{Zn}} + {m_{HCl}} - {m_{{H_2}}} = 6,5 + 220 - 0,2 = 226,3g\\
\to C{\% _{{\rm{dd}}ZnC{l_2}}} = \dfrac{{13,6}}{{226,3}} \times 100\% = 6\%
\end{array}\)
Câu 8:
\(\begin{array}{l}
{m_{{\rm{dd}}}} = {m_{NaCl}} + {m_{{H_2}O}}\\
\to {m_{{H_2}O}} = 60 - 30 = 30g\\
C\% NaCl = \dfrac{{{m_{NaCl}}}}{{{m_{{\rm{dd}}}}}} \times 100\% = \dfrac{{30}}{{60}} \times 100\% = 50\% \\
D = \dfrac{m}{V} \to V = \dfrac{m}{D} = \dfrac{{30}}{{1,17}} = 25,64ml\\
{n_{NaCl}} = 0,5mol\\
\to C{M_{NaCl}} = \dfrac{n}{V} = 19,5M
\end{array}\)
