Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{H_2SO_4}=\dfrac{62.60\%}{98}=0,38\left(mol\right)\)
PTHH:
Mg + H2SO4 ---> MgSO4 + H2
a a a a
Zn + H2SO4 ---. ZnSO4 + H2
b b b b
hệ pt \(\left\{{}\begin{matrix}24a+65b=15,4\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Zn}=0,2.65=13\left(g\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,4}{15,4}=15,58\%\\\%m_{Zn}=100\%-15,58\%=84,42\%\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}m_{MgSO_4}=0,1.120=12\left(g\right)\\m_{ZnSO_4}=0,2.161=32,2\left(g\right)\\m_{dd}=62+15,4-0,3.2=76,8\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,38-0,1-0,2\right).98=7,84\left(g\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{12}{76,8}=15,625\%\\C\%_{ZnSO_4}=\dfrac{32,2}{76,8}=41,2\%\\C\%_{H_2SO_4}=10,2\%\end{matrix}\right.\)
