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Solid iron metal reacts with oxygen gas to give iron (III) oxide according to the following unbalanced chemical equation: ______Fe (s) + ______O2 (g) ® _______ Fe2O3 (s) If you react 42.88 g of iron with excess O2, how many grams of iron (III) oxide should form?

I have balanced the equation and have 4 Fe + 3 O2 ==> 2Fe2O3
I also have calculated the molar mass of iron and oxygen..
Fe= 4 x 55.85 = 223.4 g/mol
O = 6 x 16.00 = 96 g/mol
total = 223.4 +96 = 314.4 g/mol

I then took the initial # provided in the equation and divided it by mass number of Fe (42.88 g/mol / 223.4 g/mol) multiplied by the coefficients 6mol O2 divided by coefficient of 4 mol Fe and finally divided by total molar mass of Fe2O3 319.4 g/mol.
I got a very big number (877.9g/mol Fe2O3) so I feel like I didn't solve it correctly, can someone check for me please?

Respuesta :

sebassandin

This problem is providing a chemical reaction between iron and oxygen to produce iron(III) oxide as follows:

[tex]4 Fe + 3 O_2 \rightarrow 2Fe_2O_3[/tex]

And asks for the produced mass of the oxide when 42.88 g of iron are consumed in excess oxygen. Thus, we can firstly write a calculation pathway from grams Fe to grams Fe₂O₃:

42.88 g Fe --> mol Fe --> mol Fe₂O₃ --> g Fe₂O₃

Which means we can set up the following dimensional analysis:

[tex]42.88gFe*\frac{1molFe}{55.85gFe} *\frac{2molFe_2O_3}{4molFe} *\frac{159.69gFe_2O_3}{1molFe_2O_3}\\\\= 61.30 gFe_2O_3[/tex]

This means you should no have multiplied by 6 but by 2/4 as the mole ratio of Fe to Fe₂O₃.

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