The given reaction is:
3Fe + 4H2O → Fe3O4 + 4H2
Given:
Mass of Fe = 354 g
Mass of H2O = 839 g
Calculation:
Step 1 : Find the limiting reagent
Molar mass of Fe = 56 g/mol
Molar mass of H2O = 18 g/mol
# moles of Fe = mass of Fe/molar mass Fe = 354/56 = 6.321 moles
# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles
Since moles of Fe is less than H2O; Fe is the limiting reagent.
Step 2: Calculate moles of Fe3O4 formed
As per reaction stoichiometry:
3 moles of Fe form 1 mole of Fe3O4
Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4
Step 4: calculate the mass of Fe3O4 formed
Molar mass of Fe3O4 = 232 g/mol
# moles = 2.107 moles
Mass of Fe3O4 = moles * molar mass
= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)
