\(n_{Na}=\dfrac{4,6}{23}=0,2mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,2        0,2           0,2           0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{NaOH}=0,2\cdot40=8g\)

\(m_{ddNaOH}=4,6+0,2\cdot18-0,1\cdot2=8g\)

\(\Rightarrow C\%=\dfrac{m_{NaOH}}{m_{ddNaOH}}\cdot100\%=\dfrac{8}{8}\cdot100\%=100\%???\)

Sửa đề: Tính nồng độ mol của dung dịch NaOH???

\(C_{M_{NaOH}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}M\)

vì đề không cho \(D_{NaOH}=???\) nên không tính được dung dịch NaOH

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{HCl} = 2n_{Zn} = 0,2(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{3,75\%} = 194,67(gam)\\ d) n_{ZnCl_2} = n_{Zn} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\)

giúp mik vs ạ

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

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\n_{HCl}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{ddHCl}=\dfrac{0,2\cdot36,5}{7,3\%}=100\left(g\right)\\m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\end{matrix}\right.\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

b. Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=25\%\)

=> \(m_{dd_{H_2SO_4}}=39,2\left(g\right)\)

Ta có: \(m_{H_2}=0,1.2=0,2\left(g\right)\)

=> \(m_{dd_{ZnSO_4}}=6,5+39,2-0,2=45,5\left(g\right)\)

Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

=> \(C_{\%_{ZnSO_4}}=\dfrac{16,1}{45,5}.100\%=35,4\%\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

a) PTHH : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo Pt : \(n_{Fe}=n_{H2SO4}=n_{FeSO4}=n_{H2}=0,2\left(mol\right)\)

b) \(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(C_{MddH2SO4}=\dfrac{0,2}{0,2}=1\left(M\right)\)

d) \(m_{muối}=m_{FeSO4}=0,2.152=30,4\left(g\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,2        0,4                      0,2 
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

400ml = 0,4l

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

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

        1         2              1           1

      0,1                       0,1       0,1

a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(n_{ZnCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(C_{M_{ZnCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)

 Chúc bạn học tốt

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

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

a, \(KOH+HCl\rightarrow KCl+H_2O\)

b, \(n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{KCl}=n_{KOH}=0,2\left(mol\right)\Rightarrow m_{KCl}=0,2.74,5=14,9\left(g\right)\)

c, \(n_{HCl}=n_{KOH}=0,2\left(mol\right)\Rightarrow m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{40\%}=18,25\left(g\right)\)