a) \(Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\\ MgO+H_2SO_4\xrightarrow[]{}MgSO_4+H_2O\)
b) \(n_{H_2SO_4}=\dfrac{200.24,5}{100.98}=0,5\left(mol\right)\)
Gọi x, y là số mol Mg, MgO
⇒ \(n_{H_2SO_4}=x+y=0,5\left(mol\right)\)
Ta có hpt: \(\left\{{}\begin{matrix}24x+40y=16,8\\x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
\(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
Theo BTKL: \(m_{hh}+m_{ddH_2SO_4}=m_{ddsau}+m_{H_2}\)
⇒ \(m_{ddsau}=16,8+200-0,2.2=216,4\left(g\right)\)
\(m_{MgSO_4}=120.\left(0,2+0,3\right)=60\left(g\right)\)
\(C\%_{MgSO_4}=\dfrac{60}{216,4}.100=27,73\%\)
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