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)
calculation
step 1: write a balanced chemical equation
that is 3Fe + 4 H2O → Fe3SO4 + 4H2
from the reaction above 3 moles of Fe reacted with 4 moles of water therefore Fe is the limiting reagent
Step2: find the number of mole of each reactant
that is mole= mass /molar mass
moles of Fe = 354 g/55.85 g/mol =6.338 mole
moles of H2O = 839 g/ 18 g/mol = 46.61 moles
since Fe is the limiting reagent by use of mole ratio between Fe : Fe3O4 which is 3:1 therefore the moles of Fe3O4 = 6.338 moles x 1/3=2.113 moles
mass of Fe3O4 is= moles x molar mass
= 2.113 x231.53 g/mol=489.2 grams
