When 5.00 g of iron is reacted with 5.00 g of water according to the chemical equation shown: 3 Fe(s) + 4 H2O(l) → Fe₃O₄(s) + 4 H2(g) What is the theoretical yield in grams of Fe₃O₄?

The theoretical yield can be calculated using the stoichiometry of the chemical equation. In this case, we have 3 moles of iron reacting with 4 moles of water to produce 1 mole of Fe₃O₄. The molar mass of Fe₃O₄ is calculated as 231.53 g/mol. From the given masses, we can calculate the number of moles of iron and water, and then use the mole ratio to find the moles of Fe₃O₄. Finally, we multiply the moles of Fe₃O₄ by its molar mass to find the theoretical yield in grams.

Using the given masses of 5.00 g of iron and 5.00 g of water, we can calculate the number of moles:

moles of Fe = mass of Fe / molar mass of Fe = 5.00 g / 55.85 g/mol = 0.0893 mol

moles of H₂O = mass of H₂O / molar mass of H₂O = 5.00 g / 18.02 g/mol = 0.277 mol

According to the balanced equation, the mole ratio is 3:4 between Fe and H₂O, so we can calculate the moles of Fe₃O₄:

moles of Fe₃O₄ = 0.0893 mol * (1 mol Fe₃O₄ / 3 mol Fe) = 0.0298 mol

Finally, we can find the theoretical yield in grams by multiplying the moles of Fe₃O₄ by its molar mass:

theoretical yield = moles of Fe₃O₄ * molar mass of Fe₃O₄ = 0.0298 mol * 231.53 g/mol = 6.89 g

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