For the following reaction, 89.5 grams of iron are allowed to react with 36.9 grams of oxygen gas. iron (s) oxygen (g) iron(III) oxide (s) What is the maximum amount of iron(III) oxide that can be formed

122.3 g of Fe₂O₃ is the maximum amount formed

Explanation:

Our reactants for the reaction:

89.5 g of Fe

36.9 g of O₂

We convert the mass of each to moles:

89.5 g of Fe . 1 mol / 55.85 g = 1.60 moles

36.9 g of O₂ . 1mol / 32g = 1.15 moles

The reaction is  4Fe(s) + 3O₂(g) →  2Fe₂O₃ (s)

We determine the limiting reactant:

4 moles of Iron can react with 3 moles of O₂

Therefore 1.60 moles of Iron will react with (1.60 .3) / 4 = 1.2 moles

Oxygen is the limiting reactant; we need 1.2 moles, and we only have 1.15.

Then we work with the stoichiometry again.

3 moles of oxygen can produce 2 moles of Fe₂O₃

1.15 moles of oxygen may produce (1.15 . 2) / 3 = 0.766 moles of Fe₂O₃

We convert the moles to mass: 0.766 mol . 159.7 g /1mol = 122.3 g of Fe₂O₃

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

l dont know l am s sorry