For the reaction ? fe+? h2o ⇀↽? fe3o4+? h2 , a maximum of how many grams of fe3o4 could be formed from 354 g of fe and 839 g of h2o? answer in units of g

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

2b2t

explanation:

2b2t