Step 1
The reaction provided:
3Fe (s) +4H2O (g) => Fe3O4 (s) +4H2 (g)
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Step 2
Data provided:
112.0 g of Fe and 66.7 g of water (H2O)
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Data needed:
The molar masses:
H2O) 18.0 g/mol
Fe) 55.8 g/mol
Fe3O4) 231.5 g/mol
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Step 3
The limiting reactant:
By stoichiometry:
3Fe (s) +4H2O (g) => Fe3O4 (s) +4H2 (g)
3 x 55.8 g Fe ------- 4 x 18.0 g H2O
112.0 g Fe -------- X
X = 112.0 g Fe x 4 x 18.0 g H2O/3 x 55.8 g Fe
X = 48.2 g H2O
For 112.0 g of Fe, 48.2 g of H2O is needed, but there is 66.7 g of H2O, so the limiting reactant is Fe and the excess is H2O.
Amount of water that remains:
66.7 g H2O - 48.2 g H2O = 18.5 g of H2O
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Step 4
The amount of Fe3O4 produced:
By stoichiometry,
3Fe (s) +4H2O (g) => Fe3O4 (s) +4H2 (g)
3 x 55.8 g Fe -------- 231.5 g Fe3O4
112.0 g Fe -------- X
X = 155 g Fe3O4 approx.
Answer:
Mass of excess left = 18.5 g of H2O
Mass of Fe3O4 produced = 155 g Fe3O4