a)
\(n_{hhA}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ Đặt:n_{Cl_2}=a\left(mol\right);n_{O_2}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow a+b=0,5\left(1\right)\\ Mà:m_{hhA}=m_{muối}-m_{hhB}=42,34-16,98=25,36\left(g\right)\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a+b=0,5\\71a+32b=25,36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,24\\b=0,26\end{matrix}\right.\\ \Rightarrow\%V_{Cl_2}=\dfrac{0,24}{0,5}.100=48\%\Rightarrow \%V_{O_2}=\dfrac{0,26}{0,5}.100=52\%\)
b) Đặt x,y là số mol của Mg, Al trong hhB (x,y>0) (mol)
=> 24x+27y=16,98(3)
Áp dụng ĐLBT electron:
\(n_{e.cho}=n_{e.nhận}\\ \Leftrightarrow2.n_{Mg}+3.n_{Al}=2.n_{Cl_2}+4.n_{O_2}\\ \Leftrightarrow2x+3y=2.0,24+4.0,26=1,52\left(4\right)\\ \left(3\right),\left(4\right)\Rightarrow\left\{{}\begin{matrix}2x+3y=1,52\\24x+27y=16,98\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,55\\y=0,14\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,55.24}{16,98}.100\approx77,739\%\\ \Rightarrow\%m_{Al}\approx22,261\%\)
a)\(n_{Cl_2}=x\left(mol\right),n_{O_2}=y\left(mol\right)\)
\(\Rightarrow x+y=\dfrac{11,2}{22,4}=0,5\left(1\right)\)
BTKL: \(m_A=m_{hh}-m_B=42,34-16,98=25,36g\)
\(\Rightarrow35,5\cdot2x+16\cdot2y=25,36\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}x=0,24\\y=0,26\end{matrix}\right.\)
\(\%V_{Cl_2}=\dfrac{0,24}{0,24+0,26}\cdot100\%=48\%\)
\(\%V_{O_2}=100\%-48\%=52\%\)
b)\(n_{Mg}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(\Rightarrow24a+27b=16,98\left(1\right)\)
BTe: \(2n_{Mg}+3n_{Al}=2n_{Cl_2}+4n_{O_2}\)
\(\Rightarrow2x+3y=2\cdot0,24+4\cdot0,26=1,52\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}a=0,55\\b=0,14\end{matrix}\right.\)
\(\%m_{Mg}=\dfrac{0,55\cdot24}{16,98}\cdot100\%=77,74\%\)
\(\%m_{Al}=100\%-77,74\%=22,26\%\)