Đáp án:
75ml
Giải thích các bước giải:
\(\begin{array}{l}
2Mg + {O_2} \xrightarrow{t^0} 2MgO\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
hh:Mg(a\,mol),Cu(b\,mol),Al(c\,mol)\\
BTKL:{m_{hh}} + {m_{{O_2}}} = {m_Y} \Rightarrow {m_{{O_2}}} = 3,33 - 2,13 = 1,2g\\
{n_{{O_2}}} = \dfrac{{1,2}}{{32}} = 0,0375\,mol\\
\Rightarrow 0,5a + 0,5b + 0,75c = 0,0375 \Rightarrow 2a + 2b + 3c = 0,15(1)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{MgO}} = {n_{Mg}} = a\,mol\\
{n_{CuO}} = {n_{Cu}} = b\,mol\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,5c\,mol\\
{n_{HCl}} = 2{n_{MgO}} + 2{n_{CuO}} + 6{n_{A{l_2}{O_3}}} = 2a + 2b + 3c = 0,15\,mol\\
{V_{{\rm{ddHCl}}}} = \dfrac{{0,15}}{2} = 0,075l = 75\,ml\\
\\
\end{array}\)
