Hòa tan 6,5g Zn hoàn toàn vào 200 ml dung dịch HCl
a) Viết PTHH
b) Tính VH2(đktc)
c) Tính C\(_M\)HCl phản ứng
d) Tính Khối lượng muối tạo thành
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2
0,2-->0,4-------->0,2---->0,2
\(\Rightarrow\left\{{}\begin{matrix}b,V_{H_2}=0,2.22,4=4,48\left(l\right)\\c,V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\\d,m_{MgCl_2}=0,2.95=19\left(g\right)\end{matrix}\right.\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48l\)
\(V_{HCl}=\dfrac{0,4}{2}=0,2l\)
\(m_{ZnCl_2}=0,2.95=19g\)
câu 1
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,25 0,5 0,25 0,25
\(m_{FeCl_2}=0,25.127=31,75g\\
V_{H_2}=0,25.22,4=5,6\\
C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5M\)
câu 2
1 ) \(m_{\text{dd}}=35+100=135g\\
2,C\%=\dfrac{204}{204+100}.100=60\%\\
=>m\text{dd}=\dfrac{100.204}{60}=340g\)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{H_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,25<-0,5<---0,25<--0,25
=> mZn = 0,25.65 = 16,25 (g)
c) mZnCl2 = 0,25.136 = 34 (g)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)
e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ a,m_{Fe}=0,2.56=11,2\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{HCl} = 2n_{Zn} = 0,2(mol)\\ \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{3,75\%} = 194,67(gam)\\ d) n_{ZnCl_2} = n_{Zn} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(m_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c, \(n_{HCl}=2n_{Mg}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
`a)PTHH: Zn + 2HCl -> ZnCl_2 + H_2↑`
____________________________________
`b) n_[Zn] = [ 6,5 ] / 65 = 0,1 (mol)`
Theo `PTHH` có: `n_[H_2] = n_[Zn] = 0,1 (mol)`
`-> V_[H_2 (đktc)] = 0,1 . 22,4 = 2,24 (l)`
______________________________________
`c)` Theo `PTHH` có: `n_[HCl] = 2 n_[Zn] = 2 . 0,1 = 0,2 (mol)`
Đổi `200 ml = 0,2 l`
`-> C_[M_[HCl]] = [ 0,2 ] / [0,2 ] = 1(M)`
________________________________________
`d)` Theo `PTHH` có: `n_[ZnCl_2] = n_[Zn] = 0,1 (mol)`
`-> m_[ZnCl_2] = 0,1 . 136 = 13,6 (g)`
VHCl= 200ml = 0,2 (l)
Zn + 2HCl -- > ZnCl2 + H2
nZn = 6,5 : 65 = 0,1(mol)
VH2 = 0,1 . 22,4 = 2,24 (l)
\(C_{MHCl}=\dfrac{n}{V}=\dfrac{0,2}{0,2}=1M\)
mZnCl2 = 0,1 . 136 = 13,6 (g)