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\(n_{H_2}=\dfrac{2,479}{22,4}=\dfrac{2479}{22400}mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo pt ta có: \(n_{Zn}=n_{H_2}=\dfrac{2479}{22400}mol\)\(\approx0,11mol\)
\(\Rightarrow m_{Zn}\approx7,2g\)
\(n_{HCl}=2n_{H_2}=0,22mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,22}{0,1}=2,2M\)
Mg+ 2HCl→ MgCl2+ H2
(mol) 0,1 0,2 0,1 0,1
a) \(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
→\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(lít\right)\)
b) Đổi: 100ml=0,1 lít
\(C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,2}{0,1}=2M\)
c) \(m_{MgCl_2}=n.M=0,1.95=9,5\left(g\right)\)
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
nFe = nH2 = 0,3 (mol)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b) nHCl = 2.nH2 = 0,6 (mol)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,3}=2\left(l\right)\)
c) \(n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,3}{2}=0,15M\)
`a)`
`Zn+2HCl->ZnCl_2+H_2`
`ZnO+2HCl->ZnCl_2+H_2O`
`b)`
Theo PT: `n_{Zn}=n_{H_2}={2,24}/{22,4}=0,1(mol)`
`->m_{Zn}=0,1.65=6,5(g)`
`->m_{ZnO}=14,6-6,5=8,1(g)`
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
Câu 3 :
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
1) Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
\(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{MgCl2}=0,2.136=27,2\left(g\right)\)
2) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{20}=73\left(g\right)\)
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình chỗ :
\(m_{MgCl2}=0,2.95=19\left(g\right)\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,Theo.PTHH:n_{H_2}=n_{Mg}=0,3\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,3\cdot22,4=6,72\left(l\right)\)
Ta có sau p/ứ muối tạo thành là \(MgCl_2\)
Do đó \(m=m_{MgCl_2}=0,3\cdot95=28,5\left(g\right)\)
`a) Zn + 2HCl -> ZnCl_2 + H_2`
`b) n_{Zn} = (13)/(65) = 0,2 (mol)`
Theo PT: `n_{ZnCl_2} = n_{H_2} = 0,2 (mol)`
`=> m_{ZnCl_2} = 0,2.136 = 27,2 (g)`
`c) V_{H_2} = 0,2.22,4 = 4,48 (l)`
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)