Question

When iron and steam react₄ at high temperatures, the following reaction takes place.

3 Fe(s) + 4H₂O(g) Fe₃O4 (s) +4 H₂(g)
How much iron (in g) must react with excess steam to form 817 g of Fe₃O4, if the reaction yield is 68.2%? g

Answer 1

To form 817 g of Fe₃O₄ with a reaction yield of 68.2%, approximately 265.29 g of iron must react with excess steam.

Step-by-step explanation:

To determine the amount of iron needed to form 817 g of Fe₃O₄, we need to find out the molar mass of Fe₃O₄. Fe has a molar mass of 55.845 g/mol and O has a molar mass of 15.999 g/mol.

So the molar mass of Fe₃O₄ is (3 × 55.845) + (4 × 15.999) = 231.532 g/mol.

Since the reaction yield is 68.2%, we can calculate the amount of iron needed as follows:

Amount of Fe = (mol of Fe₃O₄ × molar mass of Fe₃O₄) / reaction yield

Amount of Fe = (817 g ÷ 231.532 g/mol) / 0.682 = 4.75 mol

Now, we can calculate the mass of iron needed:

Mass of Fe = mol of Fe × molar mass of Fe = 4.75 mol × 55.845 g/mol = 265.29 g