Sửa đề thành 0,54 gam Al cho số mol đẹp bạn nhé!
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}n_{Al}=0,03\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,01\left(mol\right)\end{matrix}\right.\)
a, Ta có: \(n_{H_2}=0,03.22,4=0,672\left(l\right)\)
b, \(m_{H_2SO_4}=0,03.98=2,94\left(g\right)\)
c, \(m_{Al_2\left(SO_4\right)_3}=0,01.342=3,42\left(g\right)\)
Bạn tham khảo nhé!