Question

Cho 5,4g Al vào dung dịch H2SO4 loãng chứa 39,2g H2SO4. Tìm thể tích H2 thu được (đktc)

 

Answer 1

\(\begin{array} {l} n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \dfrac{n_{Al}}{2}<\dfrac{n_{H_2SO_4}}{3}\to H_2SO_4\text{ dư}\\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)\\ V_{H_2(đktc)}=0,3.22,4=6,72(l) \end{array}\)

Answer 2

b, Theo PT: nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)

⇒VH2=0,3.22,4=6,72(l)⇒VH2=0,3.22,4=6,72(l)