Câu 4:
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 65y = 2,11 (1)
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=x\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=y\left(mol\right)\end{matrix}\right.\)
⇒ 133,5x + 136y = 6,725 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,03\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{HCl}=3n_{Al}+2n_{Zn}=0,13\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,13.36,5}{5\%}=94,9\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Zn}=0,065\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,065.22,4=1,456\left(l\right)\)
c, \(\%m_{Zn}=\dfrac{0,02.65}{2,11}.100\%\approx61,61\%\)
d, PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
_____0,03_____0,045______________0,045 (mol)
\(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
0,02____0,02___________0,02 (mol)
→ CuSO4 dư, KL pư hết.
⇒ m chất rắn = mCu = 64.(0,045 + 0,02) = 4,16 (g)
