Fe+2HCl->Fecl2+H2
0,1----0,2----0,1---0,1
n Fe=0,1 mol
n HCl=0,3 mol
=>HCl dư
=>m FeCl2=0,1.127=12,7g
=>VH2=0,1.22,4=2,24l
=>m HCl dư=0,1.36,5=3,65g
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
\(n_{HCl}=\dfrac{10,95}{36,5}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 < 0,3 ( mol )
0,1 0,2 0,1 0,1 ( mol )
\(m_{FeCl_2}=0,1.127=12,7g\)
\(V_{H_2}=0,1.22,4=2,24l\)
\(m_{HCl\left(dư\right)}=\left(0,3-0,2\right).36,5=3,65g\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
LTL: \(0,1< \dfrac{0,3}{2}\rightarrow\) HCl dư
Theo pthh: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=0,1.2=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{HCl\left(dư\right)}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\end{matrix}\right.\)
a. \(n_{Fe}=\dfrac{5.6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{10.95}{36,5}=0,3\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,1 0,2 0,1 0,1
Ta thấy : \(\dfrac{0.1}{1}< \dfrac{0.3}{2}\) => Fe đủ , HCl dư
\(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
b. \(V_{O_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl\left(dư\right)}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\)
\(n_{Fe}=5,6:56=0,1\left(mol\right)\)
\(n_{HCl}=10,95:36,5=0,3\left(mol\right)\)
pthh : \(Fe+2HCl->FeCl_2+H_2\)
LTL :
\(\dfrac{0,1}{1}\)< \(\dfrac{0,3}{2}\)
=> HCl dư , Fe hết
theo pthh : \(n_{HCl\left(p\text{ư}\right)}=2n_{Fe}=0,2\left(mol\right)\)
=> \(n_{HCl\left(d\right)}=n_{HCl\left(b\text{đ}\right)}-n_{HCl\left(p\text{ư}\right)}=0,3-0,2=0,1\left(mol\right)\)
=> \(m_{HCl}=0,1.36,5=3,65\left(G\right)\)
theo pthh : \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(m_{FeCl_2}=0,1.127\) = 12,7 (G)
