\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
3 2 1 ( mol )
0,9 0,6 0,3 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,9.56=50,4g\)
\(V_{kk}=V_{O_2}:\dfrac{1}{5}=\left(0,6.22,4\right).5=13,44.5=67,2l\)
a. PTHH : 3Fe + 2O2 -to> Fe3O4
0,9 0,6 0,3
\(m_{Fe}=56.0,9=50,4\left(g\right)\)
b. \(V_{O_2}=0,6.22,4=13,44\left(l\right)\)
\(\Rightarrow V_{kk}=13,44.5=67,2\left(l\right)\)
\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)
a. Theo pt: \(n_{Fe}=3.n_{Fe_3O_4}=3.0,3=0,9\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,9.56=50,4\left(g\right)\)
b. Theo pt: \(n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}.0,9=0,6\left(mol\right)\)
\(\Rightarrow V_{kk}=0,6.22,4.5=67,2\left(lít\right)\)
Pt : \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4|\)
3 2 1
0,9 0,6 0,3
a) \(n_{Fe}=\dfrac{0,3.3}{1}=0,9\left(mol\right)\)
⇒ \(m_{Fe}=0,9.56=50,4\left(g\right)\)
b) \(n_{O2}=\dfrac{0,9.2}{3}=0,6\left(mol\right)\)
\(V_{O2\left(dktc\right)}=0,6.22,4=13,44\left(l\right)\)
\(V_{KK}=5.V_{O_2}=5.13,44=67,2\left(l\right)\)
Chúc bạn học tốt
