\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PT ta có tỉ lệ:
\(\dfrac{0,4}{6}>\dfrac{0,1}{3}\Rightarrow HCl_{dư}\). \(H_2\) hết nên ta tính theo \(n_{H_2}\)
a. Theo PT ta có: \(n_{Al}=\dfrac{0,1.2}{3}=0,06\left(mol\right)\)
\(\Rightarrow m_{Al}=0,06.27=1,62\left(g\right)\)
b. Theo PT ta có: \(n_{AlCl_3}=\dfrac{0,1.2}{3}=0,06\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,06.133,5=8,01\left(g\right)\)
c. Ta có: \(n_{HCl\left(pư\right)}=\dfrac{0,1.6}{3}=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)
