\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(n_{Zn}=\dfrac{65}{65}=1mol\)

\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8mol\)

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

 1    <   0,8                              ( mol )

0,4        0,8           0,4        0,4    ( mol )

\(m_{ddspứ}=200+65-0,4.2=264,2g\)

\(\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,4.136}{264,2}.100=14,93\%\\C\%_{H_2}=\dfrac{0,4.2}{264,2}.100=0,3\%\\C\%_{Zn\left(dư\right)}=\dfrac{\left(1-0,4\right).65}{264,2}.100=14,76\%\end{matrix}\right.\)

\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{17,8\%.200}{36,5}=\dfrac{356}{365}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{\dfrac{356}{365}}{2}\\ \Rightarrow Znhết,HCldư\\ n_{HCl\left(dùng\right)}=0,1.2=0,2\left(mol\right)\\ m_{HCl\left(dùng\right)}=0,2.36,5=7,3\left(g\right)\\ b.n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.n_{HCl\left(Dư\right)}=\dfrac{356}{365}-0,2=\dfrac{283}{365}\left(mol\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{6,5+200}.100\approx6,586\%\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{\dfrac{283}{365}.36,5}{6,5+200}.100\approx13,705\%\)

*Zn + 2HCl = ZnCl2 + H2
0,1.....0,2........0,1....0,1
nZn=0,1 mol
nHCl =0,3 mol
=> HCl dư
nZnCl2 = 0,1 mol => m=13,6 g
nHCl dư =0,3-0,2=0,1 mol => m=3,65 g
nH2 =0,1 mol => mH2=0,2 g
Cm(ZnCl2) = 0,1:0,2=0,5M
Cm(HCl) =0,1:0,2=0,5M
*Zn + 2HCl = ZnCl2 + H2
0,1.....0,2........0,1....0,1
nZn=0,1 mol
mHCl = 30 g => nHCl =0,82.... > 2nZn
=> HCl dư
nZnCl2 = 0,1 mol => m=13,6 g
mHCl pứ =7,3 g
=> mHCl dư = 22,7 g
nH2 =0,1 mol => mH2=0,2 g
mdd sau = 6,5+200 - 0,2=206,3 g
=> C% ZnCl2 = 6,59%
=> C% HCl = 11%

a) $Zn+ 2HCl \to ZnCl_2 + H_2$
b) $n_{HCl} = \dfrac{250.7,3\%}{36,5} = 0,5(mol)$
$n_{Zn} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol)$
$m = 0,25.65 =16,25(gam) ; V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)

$m_{dd\ sau\ pư} = 16,25 + 250 - 0,25.2 = 265,75(gam)$
$C\%_{ZnCl_2}  = \dfrac{0,25.136}{265,75}.100\% = 12,8\%$

\(a/ 4Zn+2HCl \to ZnCl_2+H_2 \\ n_{HCl}=\frac{250.7,3\%}{36,5}=0,5(mol)\\ b/ \\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\frac{1}{2}.n_{HCl}=\frac{1}{2}.0,5=0,25(mol)\\ m_{Zn}=0,25.65=16,25(g)\\ V_{H_2}=0,25.22,4=5,6(l)\\ c/ \\ C\%_{ZnCl_2}=\frac{0,25.136}{16,25+250-0,25.2}.100=12,8\% \)

a)

 
b) 

nZn=nH2=12nHCl=0,25(mol)
m=0,25.65=16,25(gam);VH2=0,25.22,4=5,6(lít)
c)

mdd sau pư=16,25+250−0,25.2=265,75(gam)

Khối lượng dịch tăng = m_X -m_H2 → Khối lượng H₂ = 2 gam

→ \(n_{H2}=1\left(mol\right)\)

Bảo toàn H có: số mol HCl = 2 mol

→ \(n_{Cl^-}=2\left(mol\right)\)

Khối lượng muối = m_X + m_Cl^- = \(m+35,5.2=m+71\left(g\right)\)

Zn+2HCl->ZnCl2+H2

0,15----------0,15-----0,15

nH2=3,36\22,4=0,15 mol

=>mZn=0,15.65=9,75g

b>

C%=0,15.136\200.100=10,2%