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Iron ore can be reduced to iron by the following reaction: Fe2O3(s) + 3H2(g) → 2Fe + 3H2O(l) (a) How many moles of Fe can be made from 1.25 moles of Fe2O3? (b) How many moles of H2 are needed to make 3.75 moles of Fe? (c) If the reaction yields 12.50 moles of H2O, what mass of Fe2O3 was used

Respuesta :

joseaaronlara

Answer:

The answer to your question is:

a) 2.5 mol of Fe

b) 5.63 mol of H2

c) 672 g of Fe2O3

Explanation:

a)                           Fe₂O₃(s)  +  3H₂(g)  →  2Fe  +  3H₂O(l)

                            1 mol of  Fe₂O₃  -------------  2 mol of Fe

                            1.25 mol             -------------    x

                            x = (1.25 x 2) / 1

                            x = 2.5 mol of Fe

b)                            Fe₂O₃(s)  +  3H₂(g)  →  2Fe  +  3H₂O(l)

                                   3 moles of H2 ----------------  2 moles of Fe

                                   x                       ----------------   3.75 mol Fe

                                  x = (3.75 x 3) / 2

                                  x = 5.63 mol of H2

c)                         Fe₂O₃(s)  +  3H₂(g)  →  2Fe  +  3H₂O(l)

                           1 mol of Fe2O3  ------------------   3 moles of H2O

                           x                          ----------------  12.5 moles of H2O

                           x = (12.5 x 1) / 3

                           x = 4.2 moles of Fe2O3

MW Fe2O3 = (2 x 56) + (16 x 3)

                   = 112 + 48

                   = 160 g

                           160 g Fe2O3 -----------------  1 mol of Fe2O3

                             x                   -----------------   4.2 moles of Fe2O3

                            x = (4.2 x 160) / 1

                           x = 672 g of Fe2O3

         

CintiaSalazar

Considering the reaction stoichiometry, you get:

(a) 2.5 moles of Fe can be made from 1.25 moles of Fe₂O₃.

(b) 4.875 moles of H₂ are needed to make 3.75 moles of Fe.

(c) 665.5699 of Fe₂O₃ are needed to make 12.50 moles of H₂O.

The balanced reaction is:

Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each product participate in the reaction:

  • Fe₂O₃: 1 mole
  • H₂: 3 moles
  • Fe: 2 moles
  • H₂O: 3 moles

(a)  You can apply the following rule of three: if by stoichiometry of the reaction 1 mole of Fe₂O₃ produces 2 moles of Fe, 1.25 moles of Fe₂O₃ produces how many moles of Fe?

[tex]amount of moles of Fe=\frac{1.25 moles of Fe_{2} O_{3} x2 moles of Fe}{1 mole of Fe_{2} O_{3} }[/tex]

amount of moles of Fe= 2.5 moles

Finally, 2.5 moles of Fe can be made from 1.25 moles of Fe₂O₃.

(b)  You can apply the following rule of three: if by reaction stoichiometry 2 moles of Fe are produced by 3 moles of H₂, 3.25 moles of Fe are produced by how many moles of H₂?

[tex]amount of moles of H_{2} =\frac{3 moles of H_{2} x3.25 moles of Fe}{2 mole of Fe }[/tex]

amount of moles of H₂= 4.875 moles

Finally, 4.875 moles of H₂ are needed to make 3.75 moles of Fe.

(c)  You can apply the following rule of three: if by reaction stoichiometry 3 moles of H₂O are produced by 1 mole of Fe₂O₃ , 12.50 moles of H₂O are produced by how many moles of Fe₂O₃?

[tex]amount of moles of Fe_{2}O_{3} =\frac{1 moles of Fe_{2} O_{3} x12.5 moles of H_{2}O}{3 moles of H_{2}O}[/tex]

amount of moles of Fe₂O₃= 4.167 moles

Being the molar mass of Fe₂O₃, that is, the mass that has 1 mole of the compound, 159.7 g/mole, the mass that participates in the reaction is calculated as:

[tex]4.167 molesx\frac{158.7 grams}{1 mole} =[/tex] 665.4699 grams

Finally, 665.5699 of Fe₂O₃ are needed to make 12.50 moles of H₂O.

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