Question

Answer 1

a. \(n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\)

PTHH : 3Fe + 2O₂ ---t^o---> Fe₃O₄

            0,75      0,5

b. \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

\(\Rightarrow V_{kk}=11,2.5=56\left(l\right)\)

Answer 2

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{42}{56}=0,75mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

0,75    0,5                          ( mol )

\(V_{O_2}=n_{O_2}.22,4=0,5.22,4=11,2l\)

\(V_{kk}=V_{O_2}.5=11,2.5=56l\)

Answer 3

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(nFe=\dfrac{42}{56}=0,75\left(mol\right)\)

\(\Rightarrow nO_2=\dfrac{2}{3}.0,75=0,5\left(mol\right)\)

\(VO_2=0,5.22,4=11,2\left(lít\right)\)

\(V_{kk}=VO_2.5=11,2.5=56\left(lit\right)\)