Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{HCl}=1.0,1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\Rightarrow Fe.dư\\ n_{H_2}=n_{FeCl_2}=n_{Fe\left(p.ứ\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ b,n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=0,1.2=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\), ta được Fe dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe\left(dư\right)}=0,1.56=5,6\left(g\right)\)
c, \(n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
\(a,n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
LTL: \(0,1>\dfrac{0,1}{2}\) => Fe dư
Theo pthh: \(n_{H_2}=n_{FeCl_2}=n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> VH2 = 0,05.22,4 = 1,12 (l)
b, Chất dư là Fe
mFe (dư) = (0,1 - 0,05).56 = 2,8 (g)
c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,1}=0,5M\)
a. Đổi 200 ml = 0,2 lít
\(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
\(n_{HCl}=2.0,2=0,2\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,1 0,2 0,1 0,1
Ta thấy : \(\dfrac{0.2}{1}>\dfrac{0.2}{2}\) => Fe dư , HCl đủ
\(m_{Fe\left(dư\right)}=\left(0,2-0,1\right).56=5,6\left(g\right)\)
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. Sau phản ứng chất tan là FeCl2
\(V_{FeCl_2}=0,1.2=0,2\left(l\right)\)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0.1}{0,2}=0,5\left(M\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,4}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, - H2SO4 dư.
\(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,2\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,2.98=19,6\left(g\right)\)
c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
nFe = 5.6/56 = 0.1 (mol)
nHCl = 0.2*2 = 0.4 (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
LTL : 0.1/1 < 0.4/2 => HCl dư
mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g)
VH2 = 0.2*22.4 = 4.48 (l)
CM FeCl2 = 0.1/0.2 = 0.5(M)
CM HCl dư = 0.2 / 0.2 = 1(M)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.
a, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
⇒ VH2 = 0,05.22,4 = 1,12 (l)
b, Sau pư, Fe dư.
Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
⇒ mFe (dư) = 0,05.56 = 2,8 (g)
c, Theo PT: nFeCl2 = nFe (pư) = 0,05 (mol)
\(\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,05}{0,1}=0,5M\)
Bạn tham khảo nhé!
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{n_{Fe}}{1}=0.1>\dfrac{n_{HCl}}{2}=\dfrac{0.1}{2}=0.05\)
\(\Rightarrow Fedư\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(m_{Fe\left(dư\right)}=\left(0.1-0.05\right)\cdot56=2.8\left(g\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0.05}{0.1}=0.5\left(M\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
PTHH :
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Trc p/u: 0,1 0,1
p/u : 0,05 0,1 0,05 0,05
Sau p/u : 0,05 0 0,05 0,05
-> Fe dư sau p/u
a) \(m_{H_2}=0,05.2=0,1\left(g\right)\)
b) sau p/ư Fe dư
\(m_{Fedư}=0,05.2,8\left(g\right)\)
c) \(m_{FeCl_2}=0,05.\left(56+35,5.2\right)=6,35\left(g\right)\)
Fe+2HCl->fecl2+H2
0,05---0,1-----------0,05
n Fe=0,1 mol
n HCl=0,1 mol
=>VH2=0,05.22,4=1,12l
b)
=>Fe dư
m Fedu=0,05.56=2,8g