Đáp án:
\(\begin{array}{l}
\% {V_{C{l_2}}} = 55,56\% \\
\% {V_{{O_2}}} = 44,44\% \\
\% {m_{C{l_2}}} = 73,5\% \\
\% {m_{{O_2}}} = 26,5\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
hh\,A:C{l_2}(a\,mol),{O_2}(b\,mol)\\
BTKL:{m_A} + {m_{Mg}} + {m_{Al}} = {m_{hh}}\\
\Rightarrow {m_A} = 37,05 - 4,8 - 8,1 = 24,15g\\
\Rightarrow 71a + 32b = 24,15(1)\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2\,mol\\
{n_{Al}} = \dfrac{{8,1}}{{27}} = 0,3\,mol\\
BT\,e:2a + 4b = 0,2 \times 2 + 0,3 \times 3 \Leftrightarrow 2a + 4b = 1,3(2)\\
(1),(2) \Rightarrow a = 0,25;b = 0,2\\
\% {V_{C{l_2}}} = \dfrac{{0,25}}{{0,25 + 0,2}} \times 100\% = 55,56\% \\
\% {V_{{O_2}}} = 100 - 55,56 = 44,44\% \\
\% {m_{C{l_2}}} = \dfrac{{0,25 \times 71}}{{24,15}} \times 100\% = 73,5\% \\
\% {m_{{O_2}}} = 100 - 73,5 = 26,5\%
\end{array}\)
