\(n_{Al}=\dfrac{081}{27}=0.03\left(mol\right)\)
\(n_{HCl}=\dfrac{2.19}{36.5}=0.06\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(2.........6\)
\(0.03.......0.06\)
\(LTL:\dfrac{0.03}{2}>\dfrac{0.06}{6}\Rightarrow Aldư\)
\(m_{Al\left(dư\right)}=\left(0.03-0.02\right)\cdot27=0.27\left(g\right)\)
\(m_{AlCl_3}=0.02\cdot133.5=2.67\left(g\right)\)
\(m_{H_2}=0.03\cdot2=0.06\left(g\right)\)
a, Ta có: \(n_{Al}=\dfrac{0,81}{27}=0,03\left(mol\right)\)
\(n_{HCl}=\dfrac{2,19}{36,5}=0,06\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Xét tỉ lệ: \(\dfrac{0,03}{2}>\dfrac{0,06}{6}\), ta được Al dư.
Theo PT: \(n_{Al\left(pư\right)}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\)
\(\Rightarrow n_{Al\left(dư\right)}=0,01\left(mol\right)\)
\(\Rightarrow m_{Al\left(dư\right)}=0,01.27=0,27\left(g\right)\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,02\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{HCl}=0,03\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{AlCl_3}=0,02.133,5=2,67\left(g\right)\)
\(m_{H_2}=0,03.2=0,06\left(g\right)\)
mAl (dư) = 0,27 (g)
Bạn tham khảo nhé!
