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For the reaction represented by the equation 2fe + o2 → 2feo, how many grams of iron(ii) oxide are produced from 8.00 mol of iron in an excess of oxygen?

Accepted Answer

Here's how to solve this problem:
1. Balanced Equation: You have the balanced equation: 2Fe + O2 → 2FeO
2. Molar Ratio: The equation shows that 2 moles of Fe react to produce 2 moles of FeO. This is a 1:1 mole ratio.
3. Grams of FeO:

Convert moles of Fe to moles of FeO using the 1:1 mole ratio: 8.00 mol Fe
(1 mol FeO / 1 mol Fe) = 8.00 mol FeO

Calculate the molar mass of FeO: Fe (55.845 g/mol) + O (15.999 g/mol) = 71.844 g/mol

Convert moles of FeO to grams: 8.00 mol FeO
(71.844 g FeO / 1 mol FeO) = 574.75 g FeO
Therefore, 574.75 grams of iron(II) oxide are produced from 8.00 moles of iron in an excess of oxygen.

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