Áp dụng ĐLBTKL, ta có:
\(m_{O_2}=36,1-23,3=12,8g\)
\(\rightarrow V_{O_2}=\dfrac{12,8}{32}.22,4=8,96l\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
x 1/2 x x ( mol )
\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)
y 1/2 y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+65y=23,3\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,7\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,7.24=16,8g\\m_{Zn}=0,1.65=6,5g\end{matrix}\right.\)
3)
2Mg+O2-to>2MgO
x-------1\2x----------x
2Zn+O2-to>2ZnO
y-------1\2y------------y
ta có \(\left\{{}\begin{matrix}24x+65y=23,3\\40x+81y=36,1\end{matrix}\right.\)
=>x=0,7 mol, y=0,1 mol
=>VO2=0,4.22,4=8,96l
=>m Mg=24.0,7=16,8g
=>m Zn=65.01,=6,5g
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