PTHH: 2Na+Cl₂\(\rightarrow\)2NaCl

n_NaCl=\(\dfrac{8,775}{58.5}=0,15\left(mol\right)\)

nếu hiệu suất là 100% thì n_NaCl=0,15/\(\dfrac{75}{100}\)=0,2(mol)

theo PTHH: n_Na=n_NaCl=0,2

\(\rightarrow\)m_Na=0,2.23=4,6(g)

theo PTHH: ⁿCl₂=\(\dfrac{1}{2}\)n_NaCl=0,1

_{\(\rightarrow\)}^VCl₂=0,1.22,4=2,24(l)

2Na +Cl2 --> 2NaCl

nNaCl=8,775/58,5=0,15(mol)

theo PTHH : nCl2=1/2nNaCl=0,075(mol)

mà H=75% =>nCl2(thực tế)=0,075/75.100=0,1(mol)

=>mCl2=0,1.71=7,1(g)

nNa=nNaCl=0,15(mol)

mà H=75%=>nNa(thực tế )=0,15/75.100=0,2(mol)

=>mNa=0,2.23=4,6(g)

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)

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

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)

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

Ta có: \(n_{ZnCl_2}=\dfrac{27,2}{136}=0,2\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Theo PT: \(n_{Zn\left(LT\right)}=n_{ZnCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn\left(LT\right)}=0,2.65=13\left(g\right)\)

Mà: H = 75% \(\Rightarrow m_{Zn\left(TT\right)}=\dfrac{13}{75\%}=\dfrac{52}{3}\left(g\right)\)

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!

a. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4}=0,6mol\)

\(\rightarrow n_{O_2}=\frac{1}{2}n_{KMnO_4}=0,3mol\)

\(\rightarrow V_{O_2}=6,72l\)

\(V_{O_2\text{thực}}=\frac{6,72.75}{100}=5,04l\)

b. \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(n_{O_2}=1,5mol\)

\(\rightarrow n_{KMnO_4}=2n_{O_2}=3mol\)

\(\rightarrow m_{KMnO_4\text{cần}}=\frac{474.100}{80}=592,5g\)

Đáp án C