Đổi 400ml = 0,4l
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2 (1)
\(n_{Al}=\frac{5,4}{27}=0,2mol\)
Theo PTHH (1) \(n_{HCl}=3n_{Al}=3.0,2=0,6mol\)
→ CM ddHCl = 0,6/0,4 = 1,5M
Theo PTHH (1) nH2 = 3/2nAl = 3/2.0,2 = 0,3(mol)
nCuO = 32/80 = 0,4(mol)
PTHH: CuO + H2 —t°-> Cu + H2O
Trước pư: 0,4 0,3(mol)
Khi pư: 0,3 0,3 0,3(mol)
Sau pư: 0,1 0 0,3(mol)
mCuO dư = 0,1. 80 = 8(g)
mCu = 0,3. 64 = 19,2(g)
Trong m có 8gCuO dư và 19,2g Cu
%CuO = (8/27,2).100% = 29,4%; %Cu = 70,6%
a) \(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PTHH: \(n_{Al}:n_{HCl}=2:6\)
\(\Rightarrow n_{HCl}=n_{Al}.3=0,2.3=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\frac{0,6}{0,4}=1,5\left(M\right)\)
b) Theo PTHH: \(n_{Al}:n_{H_2}=2:3\)
\(\Rightarrow n_{H_2}=n_{Al}.\frac{3}{2}=0,2.\frac{3}{2}=0,3\left(mol\right)\)
\(n_{CuO}=\frac{32}{80}=0,4\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(\left\{{}\begin{matrix}\frac{n_{H_2}}{1}=\frac{0,3}{1}=0,3\\\frac{n_{CuO}}{1}=\frac{0,4}{1}=0,4\end{matrix}\right.\) \(\Rightarrow\) CuO dư, H2 phản ứng hết như vậy tính toán theo \(n_{H_2}\)
\(\Rightarrow n_{CuO\left(dư\right)}=n_{CuO\left(bđ\right)}-n_{CuO\left(pứ\right)}=0,4-0,3 =0,1\left(mol\right)\)
\(\Rightarrow\%m_{Cu}=\frac{0,3.64}{0,3.64+0,1.80}.100\%=70,59\%\)
\(\Rightarrow\%m_{CuO\left(dư\right)}=24,41\%\)