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![](https://rs.olm.vn/images/avt/0.png?1311)
\(m_{HCl}=50.7,3\%=3,65\left(g\right)\\ n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ m_{Zn}=0,05.65=3,25\left(g\right)\\ V_{H_2\left(ĐKTC\right)}=0,05.22,4=1,12\left(l\right)\\ m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)
mHCl=50.7,3%=3,65(g) -> nHCl=0,1(mol)
a) PTHH: Zn + 2 HCl -> ZnCl2 + H2
nH2=nZnCl2=nZn=nHCl/2= 0,1/2=0,05(mol)
b) m=mZn=0,05.65=3,25(g)
c) V(H2,đktc)=0,05.22,4=1,12(l)
d) mZnCl2= 136.0,05= 7,8(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m=m_{Zn}=0,1.65=6,5\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ d,m_{ddZnCl_2}=6,5+100-0,1.2=106,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{106,3}.100\approx12,794\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
Zn + 2HCl → ZnCl2 + H2
nHCl = 50.7,3% = 3,65 gam
<=> nHCl = 3,65 : 36,5 = 0,1 mol
Theo tỉ lệ phản ứng => nZn = 1/2nHCl = 0,05 mol
<=> mZn = 0,05.65 = 3,25 gam.
b) nH2 = 1/2nHCl = 0,05 mol
=> VH2 = 0,05 . 22,4 = 1,12 lít.
c) m dung dịch sau phản ứng = mZn + m dd HCl - mH2 = 3,25 + 50 - 0,05.2 = 53,15 gam
mZnCl2 = 0,05.136 = 6,8 gam
<=> C% ZnCl2 = \(\dfrac{6,8}{53,15}.100\%\) = 12,8%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4\cdot36,5=14,6g\)
\(a=m_{ddHCl}=\dfrac{14,6}{14,6\%}\cdot100\%=100g\)
\(V_{H_2}=0,2\cdot22,4=4,48l\)
\(m_{ZnCl_2}=0,2\cdot136=27,2g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{HCl} = 2n_{Zn} = 0,2(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{3,75\%} = 194,67(gam)\\ d) n_{ZnCl_2} = n_{Zn} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Lập công thức hóa học của muối kẽm clorua: Z n C l 2
b) Khối lượng muối Z n C l 2 = 6,5 + 7,3 – 0,2 = 13,6 (gam)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a) PTHH:`
`Zn + 2 HCl -> ZnCl_2 + H_2`
`0,05` `0,1` `0,05` `(mol)`
`n_[HCl] = [ [ 7,3 ] / 100 . 50 ] / [ 36,5 ] = 0,1 (mol)`
`b) V_[H_2] = 0,05 . 22,4 = 1,12 (l)`
`c) m_[Zn] = 0,05 . 65 = 3,25 (g)`
\(m_{HCl}=\dfrac{50.7,3}{100}=3,65g\\
n_{HCL}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,05 0,1 0,05
\(V_{H_2}=0,5.22,4=1,12l\\
m_{Zn}=0,05.65=3,25g\)
\(a)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
\(b)\)
\(m_{HCl}=50.7,3\%=3,65g\)
\(n_{HCl}=\frac{3,65}{36,5}=0,1mol\)
\(\text{Theo phương trình}\)\(n_{Zn}=0,05mol\)
\(m_{Zn}=0,05.65=3,25g\)
\(c)\)
\(\text{Theo phương trình}\)\(n_{H_2}=0,05mol\)
\(\rightarrow V_{H_2}_{\left(đktc\right)}=0,05.22,4=1,12l\)
\(d)\)
\(\text{Theo phương trình}\)\(n_{ZnCl_2}=0,05mol\)
\(\rightarrow m_{ZnCl_2}=0,05.136=6,8g\)