Question

5,4g hỗn hợp (Mg,Al) tác dụng với dung dịch H2SO4 20% thu được 5,04 lít (đktc)
a)PTHH
b) %m Mg, %m Al
c) m dung dịch H2SO4 20%

Answer 1

\(Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\\ a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=5,4\\1,5a+b=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{9}{220}\\b=\dfrac{9}{55}\end{matrix}\right.\\ b,\%m_{Mg}=\dfrac{\dfrac{9}{55}.24}{5,4}.100\approx72,727\%\\ \Rightarrow\%m_{Al}\approx27,273\%\\ c,m_{ddH_2SO_4}=\dfrac{98.0,225.100}{20}=110,25\left(g\right)\)