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a) Fe + 2HCl -> FeCl2 + H2 (1)
b) nFe = 13,5 : 56 = 0,241 mol
Từ pt(1) => nH2 = nFe = 0,241 mol
Thể tích khí H2 là : VH2=0,241 . 22,4 = 5,3984 l
c) Từ pt(1) => nFeCl2 = nFe = 0,241 mol
=> mFeCl2 = 0,241 . 217 = 30,607g
.
a,Mg+2HCl=>MgCl2+H2
b,nHCl=0,05.3=0,15(mol)
nMg=12/24=0,5(mol)=>Mg dư, tính thao HCl
nH2=1/2 nHCl=0,075(mol)
=>VH2=0,075.22,4=1,68(l)
c,nMgCl2=nH2=0,075(mol)
mMgCl2=0,075.95=7,125(g)
a)\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b) \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
\(n_{HCl}=0,05.3=0,15\)
Ta có \(\dfrac{0,15}{2}< \dfrac{0,5}{1}\)nên Mg dư, tính theo HCl
\(n_{H_2}=\dfrac{n_{HCl}}{2}=0,075\left(mol\right)\)
\(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
c) \(n_{MgCl_2}=\dfrac{n_{HCl}}{2}=0,075\left(mol\right)\)
\(m_{MgCl_2}=0,075.95=7,125g\)
Ta có: \(n_{Fe}=\dfrac{2,24}{56}=0,04\left(mol\right)\)
a. PTHH: Fe + 2HCl ---> FeCl2 + H2
b. Theo PT: \(n_{FeCl_2}=n_{H_2}=n_{Fe}=0,04\left(mol\right)\)
=> \(m_{FeCl_2}=0,04.127=5,08\left(g\right)\)
=> \(V_{H_2}=0,04.22,4=0,896\left(lít\right)\)
c. Theo PT: \(n_{HCl}=2.n_{Fe}=2.0,04=0,08\left(mol\right)\)
=> \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)
Ta có: \(C_{\%_{HCl}}=\dfrac{2,92}{m_{dd_{HCl}}}.100\%=5\%\)
=> \(m_{dd_{HCl}}=58,4\left(g\right)\)
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\(a,PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ b,n_{FeCl_2}=n_{H_2}=n_{Fe}=0,4\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ c,m_{FeCl_2}=127.0,4=50,8\left(g\right)\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)
a) PTHH : \(Fe+2HCl-->FeCl_2+H_2\)
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH : nH2 = nFe = 0,1 (mol)
=> VH2 = \(0,1.22,4=2,24\left(l\right)\)
c) Theo PTHH : \(n_{HCl\left(pu\right)}=2n_{Fe}=0,2\left(mol\right)\)
=> mHCl = 0,2.36,5 = 7,3 (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, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)
\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)
a) Fe + 2HCl -> FeCl2 + H2 ( 1 )
b) nFe = 13,5 : 56 = 0,241 mol
Từ pt(1) => nH2 = nFe = 0,241 mol
=> VH2= 0,241 . 22,4 = 5,3984 l
c) Từ pt(1) => nFeCl2 = nH2 = 0,241 mol
=> mFeCl2 = 0,241 . 127 = 30,607g