Question

cho 2.7g nhôm vào 400ml dung dịch HCL 1M. Tính thể tích khí sinh ra ở ĐKTC

Answer 1

\(n_{Al}=0,1\left(mol\right);n_{HCl}=0,4\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ LTL:\dfrac{0,1}{2}< \dfrac{0,4}{6}\\ \Rightarrow HCldưsaupứ\\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

Answer 2

\(n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ n_{HCl}=1.0,4=0,4(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ LTL:\dfrac{0,1}{2}<\dfrac{0,4}{6}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{H_2}=0,15(mol)\\ \Rightarrow V_{H_2}=0,15.22,4=3,36(l)\)

Answer 3

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{HCl}=0,4.1=0,4\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

______0,1-------------------------->0,15

=> V_H2 = 0,15.22,4 = 3,36(l)