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\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,5.1=0,5\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{2}\), ta được HCl dư.
Theo PT: \(n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)
b, \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
c, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
d, \(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)
\(a.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl}=2n_{Fe}=0,2mol\\ m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95\%\\ b)n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1mol\\ m_{FeCl_2}=0,1.12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$FeO + 2HCl \to FeCl_2 + H_2O$
b)
Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
$m_{FeO} = 12 - 8,4 = 3,6(gam)$
$n_{FeO} =0,05(mol)$
Theo PTHH : $n_{HCl} = 2n_{Fe} + 2n_{FeO} = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)$
c) $Fe + CuSO_4 \to FeSO_4 + Cu$
$n_{Cu} = n_{Fe} = 0,15(mol) \Rightarrow m_{chất\ rắn} = m_{FeO} + m_{Cu}$
$= 3,6 + 0,15.64 = 13,2(gam)$
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\
n_{HCl}=0,5.1=0,5mol\\
Fe+2HCl\rightarrow FeCl_2+H_2\\
\Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\
Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)
\(C_{\%HCl\left(bđ\right)}=\dfrac{0,5.36,5}{200}\cdot100=9,125\%\)
a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)=n_{FeCl_2}\)
\(\Leftrightarrow m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)
b: \(n_{HCl}=2\cdot n_{FeCl_2}=2\cdot0.1=0.2\left(mol\right)\)
\(\Leftrightarrow n_{H_2}=0.1\left(mol\right)\)
\(\Leftrightarrow V_{H_2}=2.24\left(lít\right)\)
ac,nFeCl2=nFe=5,6(mol)
⇒mFeCl2=5,6⋅127=711,2(g)
bSố mol của khí hidro là: nH2=mH2/MH2=5,6/2=2,8 (mol)
Thể tích khí hidro (ở đktc) là:VH2=nH2x22,9=2,8x22,9=64,12 (lít)
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) \(n_{FeCl_2}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M\left(FeCl_2\right)}=\dfrac{0,3}{2}=0,15M\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ a,n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ n_{Fe}=n_{FeCl_2}=n_{H_2}=0,3mol\\ m_{Fe}=0,3.56=16,8g\\ b,n_{HCl}=2n_{H_2}=0,3.2=0,6mol\\ V_{ddHCl}=\dfrac{0,6}{0,3}=2l\\ c,C_{M_{FeCl_2}}=\dfrac{0,3}{2}=0,15M\)
\(a.Fe+2HCl\rightarrow FeCl_2+H_2\)
\(b.n_{Fe}=\frac{m}{M}=\frac{11.2}{56}=0.2\left(mol\right)\)
Thế số mol Fe vào phương trình ta được \(n_{HCl}=0.4mol\)
\(\Rightarrow V_{HCl}=\frac{n_{HCl}}{C_{M_{HCl}}}=\frac{0.4}{1}=0.4l\)
\(c.n_{FeCl_2}=0.2mol\Rightarrow m_{FeCl_2}=n.M=0.2\cdot127=25.4g\)
\(V_{H_2}=n_{H_2}\cdot22.4=0.2\cdot22.4=4.48l\)