CuO + 2HCl → CuCl2 + H2O
\(n_{CuO}=\frac{8}{80}=0,1\left(mol\right)\)
\(m_{HCl}=125\times20\%=25\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{25}{36,5}=\frac{50}{73}\left(mol\right)\)
Theo PT: \(n_{CuO}=\frac{1}{2}n_{HCl}\)
Theo bài: \(n_{CuO}=\frac{73}{500}n_{HCl}\)
Vì \(\frac{73}{500}< \frac{1}{2}\) ⇒ HCl dư
DD sau pư: HCl dư và CuCl2
Ta có: \(m_{dd}saupư=8+125=133\left(g\right)\)
Theo Pt: \(n_{HCl}pư=2n_{CuO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=\frac{50}{73}-0,2=\frac{177}{365}\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=\frac{177}{365}\times36,5=17,7\left(g\right)\)
\(\Rightarrow C\%_{HCl}dư=\frac{17,7}{133}\times100\%=13,31\%\)
Theo pT; \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,1\times135=13,5\left(g\right)\)
\(\Rightarrow C\%_{CuCl_2}=\frac{13,5}{133}\times100\%=10,15\%\)
CuO + 2HCl -----> CuCl2 + H2O
n\(_{CuO}\)=\(\frac{8}{80}\)=0,1 mol
m\(_{Hcl}\)=\(\frac{125.20}{100}\)=25 g
n\(_{HCl}\)=\(\frac{25}{36,5}\)=0,68 mol
=> HCl dư
dung dịch sau pư là HCl dư và CuCl\(_2\)
theo pthh: n\(_{HCl}\)=2n\(_{CuO}\)= 0,2 mol
n\(_{HCl}\)\(_{HCl}\)dư=0,68-0,2=0,48 mol
m\(_{HCl}dư\)=0,48.36,5=17,52 g
C% HCl=\(\frac{17,52}{125}.100\%\)=14%
theo pthh:n\(_{CuCl_2}\)=n\(CuO=0,1mol\)
m\(_{CuCl_2}\)=0,1.135=13,5 g
C% CuCl\(_2\)=\(\frac{13,5}{125}.100\%=10.8\%\)
CuO + 2HCL ➜ CuCL2 + H2O
T/có: nCuO = 8/80 = 0,1(mol)
Trong 125g đ HCl 20% có:
mHCL= 125 . 20% = 25(g)
⇒ nHCL= 25/36,5 = \(\frac{50}{73}\)(mol)
Vì \(\frac{0,1}{1}\)< \(\frac{50}{73}\)/ 2 nên HCL dư
Vậy dd sau PƯ gôm có CuCL2 và HCL còn dư sau PƯ
Theo PTHH ta có: nCuCL\(_2\)= nCuO= 0,1 mol
nHCl (PƯ)= 2nCuO= 2.0,1=0,2(mol)
⇒ nHCL (dư)= \(\frac{50}{73}\)- 0,2 = \(\frac{177}{365}\)(mol)
⇒ mCuCL\(_2\)= 0,1.135=13,5(g)
mHCL(dư)= \(\frac{177}{365}\). 36,5=17,7(g)
Áp dụng BTKL ta có:
m\(_{d_{ }d}\)= 8+125=133(g)
Vậy C% dd HCL dư = \(\frac{17,7}{133}\) . 100(%)=13,3%
C% dd CuCL2 = \(\frac{13,5}{133}\).100(%)= 10,15%
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