PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
a) nFe = \(\frac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: n\(H_2\) = nFe = 0,2 (mol)
=> V\(H_2\) = 0,2.22,4 = 4,48 (l)
b) Theo PT: nHCl = 2nFe = 2.0,2 = 0,4 (mol)
=> mHCl = 0,4.36,5 = 14,6 (g)
=> C%HCl = \(\frac{14,6}{200}.100=7,3\%\)
c) Theo PT: n\(FeCl_2\) = nFe = 0,2 (mol)
=> m\(FeCl_2\) = 0,2.127 = 25,4 (g)
=> m\(H_2\) = 0,2.2 = 0,4 (g)
=> mdd sau pứ = 11,2 + 200 - 0,4 = 210,8 (g)
=> C%\(FeCl_2\) = \(\frac{25,4}{210,8}.100=12,05\%\)
Fe + 2HCl → FeCl2 + H2
\(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
a) Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
b) Theo PT: \(n_{HCl}=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\frac{14,6}{200}\times100\%=7,3\%\)
c) Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)
\(m_{H_2}=0,2\times2=0,4\left(g\right)\)
\(m_{dd}saupư=m_{Fe}+m_{ddHCl}-m_{H_2}=11,2+200-0,4=210,8\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\frac{25,4}{210,8}\times100\%\approx12,05\%\)
nFe = 11,2 / 56 = 0,2 (mol)
a, PTHH :
Fe + 2HCl → FeCl2 + H2
(mol) 0,2...................0,4....................0,2............................0.2
V\(_{H_2}\) = 0,2 * 22,4 = 4,48 (lít)
b, mHCl = 0,4 * 36,5 = 14,6 (gam)
C%(HCl) = \(\frac{mHCl}{m\left(ddHCl\right)}\cdot100\%\) = \(\frac{14,6}{200}\cdot100\%\) = 7,3 (%)
c, mH2 = 0,2 * 2 = 0,4 (gam)
m(ddsau) = mFe + m(ddHCl) - mH2
= 11,2 + 200 - 0,4
= 210, 8 (gam)
Dung dịch sau pư chỉ có : 0,2 (mol) FeCl2
C%(FeCl2) = \(\frac{m_{FeCl2}}{m\left(ddsau\right)}\cdot100\%\) = \(\frac{0,2\cdot127}{210,8}\cdot100\%\) ~ 12,05 (%)
nFe= 11.2/56=0.2 mol
Fe + 2HCl --> FeCl2 + H2
0.2___0.4_____0.2____0.2
VH2=0.2*22.4=4.48l
mHCl= 0.4*36.5=14.6g
C%HCl = 14.6*100/200=7.3%
mdd sau phản ứng= 11.2+200-0.2*2=210.8g
C%FeCl2= 0.2*127/210.8*100%= 12.05%