ta có: nH2= \(\dfrac{3,36}{22,4}\)= 0,15( mol)
PTPU
Fe+ 2HCl\(\rightarrow\) FeCl2+ H2\(\uparrow\)
0,15.....0,3.........0,15.....0,15 mol
\(\Rightarrow\) C%HCl= \(\dfrac{0,15.36,5}{500}\). 100%= 1,095%
ta có: mdd sau pư= mFe+ mdd HCl- mH2
= 0,15.56+ 500- 0,15.2
= 508,1( g)
\(\Rightarrow\) C%FeCl2= \(\dfrac{0,15.127}{508,1}\). 100%= 3,75%
ta có : pthh: Fe + 2HCl -> FeCl2 + H2
số mol phân tử hidro là:
nH2 = 3.36 : 22.4 = 0.15 (mol)
theo pthh :nHcl =2nH2 = 0.3 (mol)
klg ddHcl là : 0.3*36.5 = 10.95 (g)
C% = 10.95 / 500*100%=2.19%
theo pthh nFeCl2= nH2 = 0.15
=>mfeCl2=0.15*127=19.05(g)
=>C%=19.05/500*100%=3.81%
PTHH: Fe + 2HCl → FeCl2 + H2
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a) Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{10,95}{500}\times100\%=2,19\%\)
b) Theo PT: \(n_{FeCl_2}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,15\times127=19,05\left(g\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15\times56=8,4\left(g\right)\)
\(\Rightarrow m_{ddFeCl_2}=8,4+500=508,4\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{19,05}{508,4}\times100\%=3,75\%\)
nH2 = \(\dfrac{3,36}{22,4}\) = 0,15 mol
Fe + 2 HCl -> FeCl2 + H2 \(\uparrow\)
0,15<--0,3 <----0,15<----0,15
a) C%HCl = \(\dfrac{0,3.36,5}{500}.100\%\) = 2,19 %
b) C%Muối = \(\dfrac{0,15.127}{0,15.56+500}.100\%\) = 3,747%
