\(n_{Al}=\dfrac{8.1}{27}-0.3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2..........3\)
\(0.3..........0.5\)
\(LTL:\dfrac{0.3}{2}< \dfrac{0.5}{3}\Rightarrow H_2SO_4dư\)
\(V_{H_2}=\left(\dfrac{0.3\cdot3}{2}\right)\cdot22.4=10.08\left(l\right)\)
Ta có: \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,5}{3}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
Bạn tham khảo nhé!
