a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
Dựa vào Phương trình phản ứng ta có :
\(\dfrac{n_{H_2SO_4}}{3}=\dfrac{0,4}{3}>\dfrac{0,3}{3}=\dfrac{n_{H_2}}{3}\)
Nên \(H_2SO_4\) dư
\(n_{Al}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)
\(\)\(m_{Al}=n.M=0,2.27=5,4\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)
\(m_{n_{Al_2\left(SO_4\right)_3}}=n.M=0,1.342=34,2\left(g\right)\)
\(n_{H_2SO_4}\left(phản.ứng\right)=\dfrac{0,3.3}{3}=0,3\left(mol\right)\)
\(m_{H_2SO_4}\left(phản.ứng\right)=0,3.98=29,4\left(g\right)\)
\(m_{H_2SO_4}\left(dư\right)=39,2-29,4=9,8\left(g\right)\)
