2Al + 6HCl → 2AlCl3 + 3H2
\(m_{HCl}=150\times14,6\%=21,9\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{21,9}{36,5}=0,6\left(mol\right)\)
a) Theo pT: \(n_{Al}=\frac{1}{3}n_{HCl}=\frac{1}{3}\times0,6=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
\(\Rightarrow m_{Cu}=21,4-5,4=16\left(g\right)\)
b) Theo PT: \(n_{H_2}=\frac{1}{2}n_{HCl}=\frac{1}{2}\times0,6=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3\times22,4=6,72\left(l\right)\)
c) \(m_{H_2}=0,3\times2=0,6\left(g\right)\)
Ta có: \(m_{dd}saupứ=21,4+150-0,6-16=154,8\left(g\right)\)
2Al + 6HCl -> 2AlCl3 + 3H2
0,2........0,6......................0,3
nHCl = 0,6 (mol)
mAl = 0,2.27=5,4(g)
mCu = 21,4=5,7=15,7(g)
VH2 = 0,3.22,4=6,72(mol)
mdd = 21,4+150-0,3.2=170,8(g)