PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Ta có: \(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,25\left(mol\right)\\n_{CuCl_2}=0,125\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,25\cdot36,5=9,125\left(g\right)\\C_{M_{CuCl_2}}=\dfrac{0,125}{0,5}=0,25\left(M\right)\end{matrix}\right.\)

Cảm ơn nhiều nha 🥰

Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

a, Theo PT: \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuCl_2}=0,1.135=13,5\left(g\right)\)

b, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9 : 

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

0,05--->0,1-------->0,05 

a) \(C_{MddHCl}=\dfrac{0,1}{0,1}=1\left(M\right)\)

b) \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,05}{0,1}0,5\left(M\right)\)

Câu 10 : 

\(n_{FeO}=\dfrac{3,6}{72}=0,05\left(mol\right)\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05-->0,1------->0,05

\(m_{ddHCl}=\dfrac{0,1.36,5}{10\%}100\%=36,5\left(g\right)\)

\(m_{ddspu}=3,6+36,5=40,1\left(g\right)\)

\(C\%_{FeCl2}=\dfrac{0,05.127}{40,1}.100\%=15,84\%\)

a) $n_{CuO} = \dfrac{5}{80} = 0,0625(mol)$

$CuO + 2HCl \to CuCl_2 + H_2O$

$n_{HCl} = 2n_{CuO} = 0,125(mol)$

$m_{HCl} = 0,125.36,5 = 4,5625(gam)$

b) $m_{dd\ sau\ pư} = m_{CuO} + m_{dd\ HCl} = 5 + 500 = 505(gam)$

$C\%_{CuCl_2} = \dfrac{0,0625.135}{505}.100\% = 1,67\%$

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  
          0,2       0,4           0,2 
\(m_{MgCl_2}=0,2.95=19\left(g\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

           1          2               1        1

         0,2       0,4            0,2       0,2

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

c) \(n_{ZnCl2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2........0.4..........0.2.......0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(C\%_{HCl}=\dfrac{14.6}{100}\cdot100\%=14.6\%\)

\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+100-0.2\cdot2=112.6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{27.2}{112.6}\cdot100\%=24.1\%\)

a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CO_2}=0,08\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,08}{0,2}=0,4\left(M\right)\)

c, \(n_{Na_2CO_3}=0,04\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na_2CO_3}=\dfrac{0,04.106}{10}.100\%=42,4\%\\\%m_{NaCl}=57,6\%\end{matrix}\right.\)

\(a.n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\\ n_{HCl}=2.0,1=0,2\left(mol\right)\\ m_{CuCl_2}=135.0,1=13,5\left(g\right)\\ b.m_{HCl}=0,2.36,5=7,3\left(g\right)\\ c.C_{MddHCl}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Em chưa biết làm dạng này như nào em? Vì dạng này rất cơ bản em ạ!

Cơ bản nhưng mà người mới học Hóa chưa biết làm thì sao anh ? Đâu phải ai cũng nắm nhanh kiến thức để làm :) 

nCuO=0,125mol

nH2SO4=0,5 mol

CuO +        2H2SO4-->     CuCl2 +   H2O

0,125          0,5

0,125          0,25                  0,125

C_CuCl2=0,125/0,5=0,25 M

C_{HCl du}=0,25/0,5=0,5 M

0                  0,25                  0,125

C_CuCl2= 0,125/