a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)=300\left(ml\right)\)
Đúng 2
Bình luận (0)
\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ b,m=m_{Al}=0,2.27=5,4\left(g\right)\\ c,n_{HCl}=\dfrac{6}{3}.0,3=0,6\left(mol\right)\\ V=V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)=300\left(ml\right)\)
Đúng 1
Bình luận (0)