$n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$
Gọi $n_{Cl_2} = a(mol) ; n_{O_2} = b(mol)$
$\Rightarrow m_A = 71a + 32b = 37,05 - 4,8 - 8,1(1)$
Bảo toàn electron : $2n_{Cl_2} + 4n_{O_2} = 2n_{Mg} + 3n_{Al}$
$\Rightarrow 2a + 4b = 0,2.2 + 0,3.3(2)$
Từ (1)(2) suy ra a = 0,25 ; b = 0,2
$\%m_{Cl_2} = \dfrac{0,25.71}{0,25.71 + 0,2.32}.100\% = 73,5\%$
$\%m_{O_2} = 100\% - 73,5\% = 26,5\%$
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Cl_2}=x\left(mol\right)\\n_{O_2}=y\left(mol\right)\end{matrix}\right.\)
\(m_{tăng}=m_{Cl_2}+m_{O_2}=37,05-4,8-8,1=24,15\left(g\right)\)
=> 71x + 32y = 24,15 (*)
BTe: \(2n_{Cl_2}+4n_{O_2}=3n_{Al}+2n_{Mg}=3.0,3+2.0,2=1,3\left(mol\right)\)
=> 2x + 4y = 1,3 (**)
Từ (*), (**) => x = 0,25; y = 0,2
=> \(\left\{{}\begin{matrix}\%m_{Cl_2}=\dfrac{0,25.35,5}{24,15}.100\%=36,75\%\\\%m_{O_2}=100\%-36,75\%=63,25\%\end{matrix}\right.\)
