a, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Ta có: \(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(n_{H_2}=3n_{Fe_2O_3}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, \(n_{Fe}=2n_{Fe_2O_3}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
a)n\(_{Fe_2O_3}\)=\(\dfrac{m_{Fe_2O_3}}{M_{F_2O_3}}\)=\(\dfrac{8}{160}\)=0,05(m)
PTHH : F\(_2\)O\(_3\) + 3H\(_2\) ➝ 2Fe + 3H\(_2\)O
tỉ lệ :1 3 2 3
số mol: 0,05 0,15 0,1 0,15
V\(_{H_2}\)=n\(_{H_2}\).22,4=0,15.22,4=3,36(l)
b) m\(_{Fe}\)=n\(_{Fe}\).M\(_{Fe}\)=0,1.56=5,6(g)
\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{8}{56.2+16.3}=0,05\left(mol\right)\)
\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
1 3 2 3
0,05 0,15 0,1 0,15
a) \(V_{H_2}=n.24,79=0,15.24,79=3,7185\left(l\right)\)
b) \(m_{Fe}=n.M=0,1.56=5,6\left(g\right).\)
