a: \(n_{MgO}=40\cdot m_{MgO}\)
\(n_{Cuo}=80\cdot m_{CuO}\)
=>\(\dfrac{40\cdot m_{MgO}}{80\cdot m_{CuO}}=\dfrac{1}{2}\)
=>\(\dfrac{m_{MgO}}{m_{CuO}}=1\)
=>\(m_{MgO}=m_{CuO}=\dfrac{48}{2}=24\left(g\right)\)
b: \(\Leftrightarrow\dfrac{80\cdot m_{CuO}}{40\cdot m_{MgO}}=\dfrac{2}{1}\)
\(\Leftrightarrow m_{CuO}=m_{MgO}=24\left(g\right)\)
`a)` Gọi \(\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{CuO}=2x\left(mol\right)\end{matrix}\right.\)
`=> 40x + 80.2x = 48`
`=> x = 0,24 (mol)`
`=>` \(\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,24.40}{48}.100\%=20\%\\\%m_{CuO}=100\%-20\%=80\%\end{matrix}\right.\)
`b)` Gọi \(\left\{{}\begin{matrix}n_{MgO}=2a\left(mol\right)\\n_{CuO}=a\left(mol\right)\end{matrix}\right.\)
`=> 2a.40 + 80a = 48`
`=> a = 0,3`
`=>` \(\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,3.2.40}{48}.100\%=50\%\\\%m_{CuO}=100\%-50\%=50\%\end{matrix}\right.\)