To determine the grams of steam needed to produce 381 g of Fe3O4, we need to first determine the molar mass of Fe3O4.

The molar mass of Fe3O4 can be calculated as follows:

Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of Fe3O4 = (3 * molar mass of Fe) + (4 * molar mass of O)
= (3 * 55.845 g/mol) + (4 * 16.00 g/mol)
= 167.535 g/mol + 64.00 g/mol
= 231.535 g/mol.

Now, using the balanced equation, we can determine the molar ratio between Fe3O4 and H2O:

3Fe + 4H2O -> Fe3O4 + H2

From this equation, we can see that 4 moles of H2O are required to produce 1 mole of Fe3O4.

To calculate the moles of Fe3O4 required to produce 381 g, we can use the following equation:

moles = mass / molar mass
moles = 381 g / 231.535 g/mol
moles ≈ 1.646 mol.

Since the molar ratio between Fe3O4 and H2O is 4:1, we can calculate the moles of H2O required using the following equation:

moles of H2O = 4 moles of Fe3O4 * (1 mole H2O / 1 mole Fe3O4)
moles of H2O = 1.646 mol * (4/1)
moles of H2O ≈ 6.584 mol.

Finally, to determine the grams of steam needed to produce 381 g of Fe3O4, we can use the following equation:

grams = moles * molar mass
grams = 6.584 mol * 18.015 g/mol (molar mass of H2O)
grams ≈ 118.64 g.

Therefore, the grams of steam needed to produce 381 g of Fe3O4 is approximately 118.64 g.

The correct answer is not provided among the options, as the calculated value is different from all the given options.

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