PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, Ta có: \(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,03\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,03.22,4=0,672\left(l\right)=672\left(cm^3\right)\)
⇒ Sai số: 672 - 660,8 = 11,2
b, Ta có: \(n_{H_2}=\dfrac{0,6608}{22,4}=0,0295\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{59}{3000}\left(mol\right)\)
\(\Rightarrow m_{Al}=\dfrac{59}{3000}.27=0,531\left(g\right)\)
⇒ Lượng tạp chất là: 0,54 - 0,531 = 0,009 (g)
Bạn tham khảo nhé!