PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
a________________a_____a (mol)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
b________________b____b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}24a+65b=10,1\\a+b=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{muối}=0,15\cdot120+0,1\cdot161=34,1\left(g\right)\\\%m_{Mg}=\dfrac{0,15\cdot24}{10,1}\approx35,64\%\\\%m_{Zn}=64,36\%\end{matrix}\right.\)
a,\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: x x
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: y y
Ta có:\(\left\{{}\begin{matrix}24x+65y=10,1\\x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,15 0,15
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 0,1 0,1
⇒ mmuối = 0,15.120 + 0,1.161 = 34,1 (g)
b,\(\%m_{Mg}=\dfrac{0,15.24.100\%}{10,1}=35,64;\%m_{Zn}=100-35,64=64,36\%\)
