35.0 grams of iron (Fe) is reacted, how many moles of iron are consumed?
How many moles of Fe3O4 will be produced?
How many grams of Fe3O4 will be produced?
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. . . . 35.0 g . . . . . . . . . . . . . . . . . . . . . . . . . . . 48.37 g
. . . 3Fe(s) . + . 4H2O(g) . → . 4H2(g) . + . Fe3O4(s)
(55.845g/mol) . . . . . . . . . . . . . . . . . . . . . . . (231.533g/mol)
. 0.6267 mol . . . . . . . . . . . . . . . .. . . . . . . . . 0.2089 mol
. . . .|_____________3 to 1________________|
so to 3 sig figs
mol Fe = 0.627 mol
mol Fe3O4 = 0.209 mol
mass of Fe3O4 = 48.4 g
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3 Fe + 4 H2O → 4 H2 + Fe3O4
Supposing excess water:
(35.0 g Fe) / (55.8450 g Fe/mol) = 0.627 moles of iron consumed
(0.627 mol Fe) x (1 mol Fe3O4 / 3 mol Fe) = 0.209 mol Fe3O4 produced
(0.209 mol Fe3O4) x (231.5326 g Fe3O4/mol) = 48.4 g Fe3O4 produced
Molar mass of Fe = 55.85 g
Molar mass of Fe₃O₄ = (55.8×3 + 16.0×4) g/mol = 231.4 g/mol
3Fe(s) + 4H₂O(g) → 4H₂(g) + Fe₃O₄(s)
Mole ratio Fe : Fe₃O₄ = 3 : 1
No. of moles of Fe reacted = (35.0 g) / (55.8 g/mol) = 0.627 mol
No. of moles of Fe₃O₄ produced = (0.627 mol) × (1/3) = 0.209 mol
Mass of Fe₃O₄ produced = (0.209 mol) × (231.4 g/mol) = 48.4 g
I have tried everything - each time I bring it over it takes out all the subscripts...how do I use subscripts to ask the question?
write it with the correct subscripts and I may be able to read it
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Answers & Comments
...
. . . . 35.0 g . . . . . . . . . . . . . . . . . . . . . . . . . . . 48.37 g
. . . 3Fe(s) . + . 4H2O(g) . → . 4H2(g) . + . Fe3O4(s)
(55.845g/mol) . . . . . . . . . . . . . . . . . . . . . . . (231.533g/mol)
. 0.6267 mol . . . . . . . . . . . . . . . .. . . . . . . . . 0.2089 mol
. . . .|_____________3 to 1________________|
so to 3 sig figs
mol Fe = 0.627 mol
mol Fe3O4 = 0.209 mol
mass of Fe3O4 = 48.4 g
When you get a good response,
please consider giving a best answer.
This is the only reward we get.
You may have to wait an hour to award BA.
3 Fe + 4 H2O → 4 H2 + Fe3O4
Supposing excess water:
(35.0 g Fe) / (55.8450 g Fe/mol) = 0.627 moles of iron consumed
(0.627 mol Fe) x (1 mol Fe3O4 / 3 mol Fe) = 0.209 mol Fe3O4 produced
(0.209 mol Fe3O4) x (231.5326 g Fe3O4/mol) = 48.4 g Fe3O4 produced
Molar mass of Fe = 55.85 g
Molar mass of Fe₃O₄ = (55.8×3 + 16.0×4) g/mol = 231.4 g/mol
3Fe(s) + 4H₂O(g) → 4H₂(g) + Fe₃O₄(s)
Mole ratio Fe : Fe₃O₄ = 3 : 1
No. of moles of Fe reacted = (35.0 g) / (55.8 g/mol) = 0.627 mol
No. of moles of Fe₃O₄ produced = (0.627 mol) × (1/3) = 0.209 mol
Mass of Fe₃O₄ produced = (0.209 mol) × (231.4 g/mol) = 48.4 g
I have tried everything - each time I bring it over it takes out all the subscripts...how do I use subscripts to ask the question?
write it with the correct subscripts and I may be able to read it