Question

Please help, and I’ll thank you.

10.0 g of iron is reacted with 10.0 g of water according to the balanced chemical equation shown below. Determine the amount of Fe3O4 formed. (molar mass of Fe3O4 is 231.55 g/mol) (Hint: Determine the limiting reactant first)

3 Fe(s) + 4 H2O(l) —— Fe3O4(s) + 4 H2(g)

Answer 1

There will be formed 13.82 grams of Fe3O4.

Step-by-step explanation:

Step 1: the balanced reaction

3 Fe(s) + 4 H2O(l) —— Fe3O4(s) + 4 H2(g)

We see that for 3 moles Fe consumed, there will be consumed 4 moles of H20 and produced 1 mole of Fe3O4 and 4 moles of H2.

Step 2: Calculating moles

⇒mole = mass / Molar mass

⇒moles Fe = 10g / 55.845g/mole = 0.179 moles

⇒moles H2O = 10g / 18g/mole = 0.5556 moles

Since we have 4 moles of H2O consumed for 3 moles of Fe, this means that for 0.179 moles of Fe, we have :

0.179 / 4 *3 = 0.2387 moles of H2O consumed

This means that Fe will be completely consumed and there will remain 0.2387 moles of H2O.

Fe is the limiting reactant and H2O is in excess.

Step 3: Calculate moles of Fe3O4

For 3 moles of Fe consumed, we have 1 mole of Fe3O4 produced.

This means, that for 0.179 moles of Fe consumed, we have 0.179/3 =0.05967 molesof Fe3O4 produced.

Step 4: Calculating mass of Fe3O4

mass= number of moles * Molar mass

mass Fe3O4 = 0.05967 mole * 231.55g/mole = 13.8166 g ≈ 13.82grams of Fe3O4

There will be formed 13.82 grams of Fe3O4.