One of the reactions for rusting iron is as follows: 4fe + 3o2 → 2fe2o3 (mm fe: 55.85 g/mol; mm o2=32 g/mol; mm fe2o3=159.70 g/mol) if 2.80 moles of iron, fe, rust completely, how many grams of iron (iii) oxide are formed? a. 0.0351 g b. 313 g c. 894 g d. 224 g

D. 224 g

Explanation:

Let's consider the rusting of iron reaction.

4 Fe + 3 O₂ → 2 Fe₂O₃

The molar ratio of Fe to Fe₂O₃ is 4:2. The moles of Fe₂O₃ produced from 2.80 moles of Fe are:

2.80 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) = 1.40 mol Fe₂O₃.

The molar mass of Fe₂O₃ is 159.70 g/mol. The mass of Fe₂O₃ is:

1.40 mol × (159.70 g / 1 mol) = 224 g

a compound is a molecule made of atoms from different elements.

a. by the combination of an unlimited number of elements

the answer is 0.42 is that is an option. this question provides no info