Giải thích các bước giải:
Bài 6:
\(\begin{array}{l}
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{m_{{\rm{dd}}HCl}} = V \times D = 100,8 \times 1,19 = 119,95g\\
\to {m_{HCl}} = \dfrac{{119,95 \times 36,5\% }}{{100\% }} = 43,78g\\
\to {n_{HCl}} = 1,2mol\\
\\
\end{array}\)
Gọi a và b lần lượt là số mol của Zn và ZnO
\(\begin{array}{l}
\left\{ \begin{array}{l}
65a + 81b = 42,2\\
2a + 2b = 1,2
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,4\\
b = 0,2
\end{array} \right.\\
\to {n_{Zn}} = 0,4mol \to {m_{Zn}} = 0,4 \times 65 = 26g\\
\to {n_{ZnO}} = 0,2mol \to {m_{ZnO}} = 0,2 \times 81 = 16,2g\\
\to \% {m_{Zn}} = \dfrac{{26}}{{42,2}} \times 100\% = 61,6\% \\
\to \% {m_{ZnO}} = \dfrac{{16,2}}{{42,2}} \times 100\% = 38,4\% \\
b)\\
{n_{{H_2}}} = {n_{Zn}} = 0,4mol\\
\to {V_{{H_2}}} = 0,4 \times 22,4 = 8,96l\\
c)\\
{n_{ZnC{l_2}}} = {n_{Zn}} + {n_{ZnO}} = 0,6mol\\
\to {m_{ZnC{l_2}}} = 0,6 \times 136 = 81,6g\\
\to {m_{{\rm{ddA}}}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 42,2 + 119,95 - 0,4 \times 2\\
\to {m_{{\rm{ddA}}}} = 161,35g\\
\to C{\% _{{\rm{dd}}A}} = \dfrac{{81,6}}{{161,35}} \times 100\% = 50,57\%
\end{array}\)
Bài 7:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{{H_2}}} = 0,2mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,2mol \to {m_{Fe}} = 11,2g\\
\to {m_{CuO}} = 24g \to {n_{CuO}} = 0,3mol\\
\to \% {m_{Fe}} = \dfrac{{11,2}}{{35,2}} \times 100\% = 31,82\% \\
\to \% {m_{CuO}} = 68,18\% \\
\to {n_{{H_2}S{O_4}}} = {n_{Fe}} + {n_{CuO}} = 0,5mol\\
\to {m_{{H_2}S{O_4}}} = 49g\\
\to {m_{{H_2}S{O_4}{\rm{dd}}}} = 35,2 + 800 - 0,2 \times 2 = 834,8g\\
\to C\% {H_2}S{O_4}{\rm{dd}} = \dfrac{{49}}{{834,8}} \times 100\% = 5,87\% \\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,2mol\\
\to {m_{FeS{O_4}}} = 30,4g\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,3mol\\
\to {m_{CuS{O_4}}} = 48g
\end{array}\)
