a)Fe +2HCl--->FeCl2 +H2
b) Ta có
n\(_{H2}=\frac{0,896}{22,4}=0,04\left(mol\right)\)
Theo pthh
n\(_{Fe}=n_{H2}=0,04\left(mol\right)\)
m\(_{Fe}=0,04.56=2,24\left(g\right)\)
c) Theo pthh
n\(_{FeCl2}=n_{H2}=0,04\left(mol\right)\)
m\(_{FeCl2}=0,04.127=5,08\left(g\right)\)
Chúc bạn học tốt
\(n_{H_2}=\frac{V}{22,4}=\frac{0,896}{22,4}=0,04mol\)
a)PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,04 0,08 0,04 0,04 (mol)
b)\(m_{Fe}=n.M=0,04.56=2,24g\)
c)\(m_{FeCl_2}=n.M=0,04.127=5,08g\)
a, PTHH : Fe+2HCl--->FeCl2+H2
b, NH2 = 0,04 (mol)
=>mFe = 0,04. 56 = 2,24g
c,mFeCl2 = 5,08g
