Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
Ta có nCuO = \(\dfrac{8}{80}\) = 0,1 ( mol )
CuO + 2HCl \(\rightarrow\) CuCl2 + H2
0,1........0,2...........0,1........0,1
=> mHCl = 0,2 . 36,5 = 7,3 ( gam )
=> mHCl cần dùng = 7,3 : 7,3 . 100 = 100 ( gam )
Ta có Mdung dịch = Mtham gia - MH2
= 8 + 100 - 0,1 . 2
= 107,8 ( gam )
=> mCuCl2 = 0,1 . 135 = 13,5 ( gam )
=> C%CuCl2 = \(\dfrac{13,5}{107,8}\times100\approx12,5\%\)
\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(0.02.......0.02.................0.02\)
\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)
\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)
PTHH:
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,04 0,04 0,04
\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư
\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)
PTHH:\(Na_2SO_3+CaCl_2\rightarrow2NaCl+CaSO_3\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_3}=\dfrac{265\cdot10\%}{126}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2}=\dfrac{500\cdot6,66\%}{111}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỷ số: \(\dfrac{53}{252}< \dfrac{0,3}{1}\) \(\Rightarrow\) CaCl2 còn dư, Na2SO3 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=\dfrac{53}{126}\left(mol\right)\\n_{CaSO_3}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=\dfrac{53}{126}\cdot58,5\approx24,61\left(g\right)\\m_{CaSO_3}=\dfrac{53}{252}\cdot120\approx25,24\left(g\right)\\m_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\cdot111\approx9,95\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2SO_3}+m_{ddCaCl_2}-m_{CaSO_3}=739,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{24,61}{739,76}\cdot100\%\approx3,33\%\\C\%_{CaCl_2\left(dư\right)}=\dfrac{9,95}{739,76}\cdot100\%\approx1,35\%\end{matrix}\right.\)
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\(_2\)=\(\frac{13,5}{125}.100\%=10.8\%\)