1.
mNaCl lí thuyết = \(\dfrac{4,68.100}{80}\) = 5,85 (g)
=> nNaCl lí thuyết = \(\dfrac{5,85}{58,5}\) = 0,1 (mol)
2Na + Cl2 ----> 2NaCl
0,1 0,05 0,1 (mol)
=> mNa = 0,1.23 = 2,3 (g)
=> VCl2 = 0,05.22,4 = 1,12 (l)
2.

n_Zn = \(\dfrac{19,5}{65}\)= 0,3 (mol)

n_Cl2 = \(\dfrac{7}{22,4}\)= 0,3125 (mol)

Zn + Cl₂ ----> ZnCl₂

0,3 0,3 (mol)

Tỉ lệ: \(\dfrac{0,3}{1}<\dfrac{0,3125}{1}\)=> Zn phản ứng hết, Cl₂ phản ứng dư.

=> m_{ZnCl2 lí thuyết} 0,3.136 = 40,8 (g)

=> H = \(\dfrac{36,72.100}{40,8}\)= 90%

3.

Cu + Cl₂ ----> CuCl₂

0,2 0,2 (mol)

=> m_{CuCl2 lí thuyết} = 0,2.135 = 27 (g)

=> m_{CuCl2 thực tế} = \(\dfrac{27.83}{100}\)= 22,41 (g)

PT: \(2Na+Cl_2\underrightarrow{t^o}2NaCl\)

Ta có: \(n_{NaCl}=\dfrac{4,68}{58,5}=0,08\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Na\left(LT\right)}=n_{NaCl}=0,08\left(mol\right)\\n_{Cl_2\left(LT\right)}=\dfrac{1}{2}n_{NaCl}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na\left(LT\right)}=0,08.23=1,84\left(g\right)\\V_{Cl_2\left(LT\right)}=0,04.22,4=0,896\left(l\right)\end{matrix}\right.\)

Mà: H% = 80%

\(\Rightarrow\left\{{}\begin{matrix}m_{Na\left(TT\right)}=\dfrac{1,84}{80\%}=2,3\left(g\right)\\V_{Cl_2\left(TT\right)}=\dfrac{0,896}{80\%}=1,12\left(l\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

\(n_{NaCl}=\dfrac{4,68}{58,5}=0,08\left(mol\right)\)

PTHH: 2Na + Cl₂ --t^o--> 2NaCl

           0,08<-0,04<-----0,08

 \(m_{NaCl\left(pư\right)}=0,08.23=1,84\left(g\right)\)

=> \(m_{NaCl\left(tt\right)}=\dfrac{1,84.100}{80}=2,3\left(g\right)\)

\(n_{Cl_2\left(tt\right)}=\dfrac{0,04.100}{80}=0,05\left(mol\right)\)

=> V_Cl2 = 0,05.22,4 = 1,12 (l)

Câu 1 :

PTHH : \(2NaCl+2H_2O\rightarrow2NaOH+Cl_2+O_2\)

\(n_{Cl_2}=\frac{V_{Cl_2}}{22,4}=\frac{3,36}{22,4}=0,15\left(mol\right)\)

- Theo PTHH : \(n_{NaCl}=2n_{Cl_2}=2.0,15=0,3\left(mol\right)\)

-> \(m_{NaCl}=n_{NaCl}.M_{NaCl}=0,3.\left(23+35,5\right)=17,55\left(g\right)\)

Câu 2 :

PTHH :................ \(Fe+Cl_2\rightarrow FeCl_2\)

Trước phản ứng :.0,012..0,02..........

Trong phản ứng :.0,012..0,012...........

Sau phản ứng : ........0....0,008......

-> Sau phản ứng Fe phản ứng hết, Cl₂ còn dư .

- Theo PTHH : \(n_{FeCl_2}=n_{Fe}=0,012\left(mol\right)\)

-> \(m_{FeCl_2}=n.M=0,012.\left(56+35,5.2\right)=1,524\left(g\right)\)

a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, Ta có: \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

d, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

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

\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

Bạn tham khảo nhé!

2Na+Cl2->2NaCl

nNaCl=4,68/58,5=0,08 mol

=>nNa=0,08 mol

nCl2=0,04 mol

mNa=0,08x23/80%=2,3 g

VCl2=0,04x22,4/70%=1,28 l

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = 2n_{H_2} = 0,3(mol)\ m_{HCl} = 0,3.36,5 = 10,95(gam)\)

\(n_{Fe}=\dfrac{8.4}{56}=0.15\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.15......0.3..................0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)