Question

Answer 1

\(1.pthh:Zn+2HCl--->ZnCl_2+H_2\uparrow\)

\(2.\) Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)

Vậy HCl dư.

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

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

\(3.\) Ta có: \(n_{HCl_{PỨ}}=2.n_{Zn}=2.0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl_{dư}}=\left(0,4-0,2\right).36,5=7,3\left(g\right)\)

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

\(\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

Ta có: \(m_{dd_{sau.PỨ}}=6,5+100-0,1.2=106,3\left(g\right)\)

\(\Rightarrow C_{\%_{HCl_{dư}}}=\dfrac{7,3}{106,3}.100\%=6,87\%\)

\(C_{\%_{ZnCl_2}}=\dfrac{13,6}{106,3}.100\%=12,79\%\)

Answer 2

1,

n Zn = 6.5/65 = 0.1 mol

 \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

1mol      2mol       1mol       1mol

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

V H2 = 0.1 x 22.4 = 2.24 L

C% HCl = 0.2x36.5x100/100 = 7,3%

C% ZnCl2 = 0.1 x 136x100/100 = 13,6%

 

Answer 3

9

Zn+2HCl->ZnCl₂+H₂

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

n _Zn=\(\dfrac{6,5}{65}\)=0,1 mol

=>HCl dư 

=>V_H2=0,1.22,4=2,24l

=>C%_ZnCl2=\(\dfrac{0,1.136}{6,5+100-0,1.2}\).100=12,79%

=>C%_HCldư=\(\dfrac{0,2.36,5}{6,5+100-0,1.2}.100\)=6,867%