Đáp án:
\(\begin{array}{l}
10\,C\\
11\,A\\
12\,A\\
13\,B
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
10)\\
Ca + 2HCl \to CaC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{M_{Ca}} = {M_{MgO}} = 40\,g/mol\\
{n_{hh}} = \dfrac{{20}}{{40}} = 0,5\,mol\\
{n_{HCl}} = 2{n_{hh}} = 1\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{1}{2} = 0,5l = 500ml\\
11)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
MgS{O_4} + 2NaOH \to Mg{(OH)_2} + N{a_2}S{O_4}\\
F{e_2}{(S{O_4})_3} + 6NaOH \to 3N{a_2}S{O_4} + 2Fe{(OH)_3}\\
Mg{(OH)_2} \to MgO + {H_2}O\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
hh:Mg(a\,mol),F{e_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
24a + 160b = 20\\
40a + 160b = 28
\end{array} \right.\\
\Rightarrow a = 0,5;b = 0,05\\
{n_{{H_2}}} = {n_{Mg}} = 0,5\,mol\\
{V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
12)\\
{n_{Cu}} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
{n_{S{O_2}}} = {n_{Cu}} = 0,1\,mol\\
{n_{{O_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{O_2}}} = 0,05 \times 22,4 = 1,12l\\
13)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,3\,mol\\
BTKL:\\
{m_{hh}} + {m_{{H_2}S{O_4}}} = {m_m} + {m_{{H_2}}}\\
\Rightarrow {m_m} = 14,5 + 0,3 \times 98 - 0,3 \times 2 = 43,3g
\end{array}\)
