How many grams of water and iron must react to produce 543 g of iron oxide, Fe3O4?

3Fe + 4H2O---> Fe3O4 + 4H2

You have the balanced equation.

1. Convert 543 g Fe3O4 to moles. Moles = grams/molar mass.
2. Using the coefficients in the balanced equation, convert moles Fe3O4 to moles of what you want (moles Fe).
3. Now convert moles Fe to grams.
grams = moles x molar mass.

To determine the amount of water and iron required to produce 543 g of iron oxide (Fe3O4), we need to calculate the molar mass of Fe3O4 and use stoichiometry.

Step 1: Calculate the molar mass of Fe3O4.
Fe has a molar mass of 55.85 g/mol, and O has a molar mass of 16.00 g/mol. Since there are three Fe atoms and four O atoms in Fe3O4, we multiply the molar mass of Fe by 3 and the molar mass of O by 4, and then add them together:
Molar mass of Fe3O4 = (3 * molar mass of Fe) + (4 * molar mass of O) = (3 * 55.85 g/mol) + (4 * 16.00 g/mol) = 159.70 g/mol

Step 2: Calculate the number of moles of Fe3O4.
To convert grams to moles, we use the equation:
Number of moles = Mass / Molar mass
Number of moles of Fe3O4 = 543 g / 159.70 g/mol = 3.400 mol

Step 3: Use stoichiometry to determine the number of moles of iron and water.
From the balanced chemical equation, we know that 3 moles of Fe react with 4 moles of H2O to produce 1 mole of Fe3O4.
So, the ratio of Fe to Fe3O4 is 3:1, and the ratio of H2O to Fe3O4 is 4:1.
Therefore, the number of moles of Fe required = (3.400 mol Fe3O4) * (3 mol Fe / 1 mol Fe3O4) = 10.2 mol Fe
Similarly, the number of moles of H2O required = (3.400 mol Fe3O4) * (4 mol H2O / 1 mol Fe3O4) = 13.6 mol H2O

Step 4: Calculate the mass of iron and water.
Finally, we need to convert the number of moles of iron and water to grams using their respective molar masses:
Mass of iron = number of moles of Fe * molar mass of Fe = 10.2 mol * 55.85 g/mol = 569.07 g
Mass of water = number of moles of H2O * molar mass of H2O = 13.6 mol * 18.02 g/mol = 245.05 g

Therefore, to produce 543 g of iron oxide, you would need approximately 569.07 g of iron and 245.05 g of water.