ta có :
2Mg+O2-to>2MgO
x--------0,5x
2Zn+O2-to>2ZnO
y-------0,5y
=>\(\left\{{}\begin{matrix}24x+65y=14,58\\0,5x+0,5y=0,15\end{matrix}\right.\)
=>x=0,12 mol ,y=0,18 mol
=>%mMg=\(\dfrac{0,12.24}{14,58}100\)=19,753%
=>%mZn=80,247%
\(n_{O_2}=\dfrac{4,8}{32}=0,15mol\)
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 ( mol )
\(2Zn+O_2\rightarrow\left(t^o\right)2ZnO\)
y 1/2 y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+65y=14,58\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,12\\y=0,18\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,12.24=2,88g\\m_{Zn}=0,18.65=11,7g\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{2,88}{14,58}.100=19,75\%\\\%m_{Zn}=100\%-19,75\%=80,25\%\end{matrix}\right.\)
\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)
gọi số mol Mg : a , số mol Zn : b (a,b>0)
pthh : \(2Mg+O_2\underrightarrow{t^o}2MgO\)
a \(\dfrac{1}{2}a\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
b \(\dfrac{1}{2}b\)
mà 24a + 65b=14,58
có : \(\dfrac{1}{2}a+\dfrac{1}{2}b=0,15\)
=> a = 0,12 ( mol ) , b=0,18 (mol)
=> \(\%m_{Zn}=\dfrac{0,18.65}{14,68}.100\%=79,7\%\\
\%m_{Mg}=100\%-79,7\%=20,3\%\)