Question

cho 5,6 (g) Fe+ 4,48(l)O2(đktc)-Fe2O3

a) tính khối lượng 

b) v o2

 

Answer 1

4Fe+3O2-to>2Fe2O3

0,1-----0,075-------0,05

n Fe=0,1 mol

n O2=0,2 mol

=>O2 dư

=>m Fe2O3=0,05.160=8g

=>VO2=0,075.22,4=1,68l

Answer 2

a. \(n_{Fe}=\dfrac{5.6}{56}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

PTHH : 4Fe + 3O₂ -> 2Fe₂O₃

              0,1     0,075        0,05

Ta thấy \(\dfrac{0.1}{4}< \dfrac{0.2}{3}\) => Fe đủ , O₂ dư

\(m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b. \(V_{O_2}=0,075.22,4=1,68\left(l\right)\)