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Nezavi [6.7K]
3 years ago
8

For the following reaction, 3.76 grams of iron are mixed with excess oxygen gas . The reaction yields 4.29 grams of iron(II) oxi

de . iron ( s ) oxygen ( g ) iron(II) oxide ( s ) What is the theoretical yield of iron(II) oxide
Chemistry
1 answer:
Ainat [17]3 years ago
7 0

Answer:

4.84g of FeO is the theoretical yield

Explanation:

The Iron, Fe(s), reacts with oxygen, O₂(g), producing Iron (II) oxide, as follows:

2Fe(s) + O₂(g) → 2FeO

<em>Theoretical yield is the yield of a reaction in which you assume the 100% of reactants is converted in products.</em>

<em />

To find theoretical yield we need to find moles of Iron, and, knowing 2 moles of Fe produce 2 moles of FeO (Ratio 1:1), we can find theoretical yield of FeO as follows:

<em>Moles Fe (Molar mass: 55.845g/mol)</em>

Using the molar mass of the compound we can convert grams to moles, thus:

3.76g Fe × (1mol / 55.845g) = <em>0.0673 moles of Fe</em>

<em />

<em>Moles and mass of FeO</em>

As there are in reaction 0.0673 moles Fe, assuming a theoretical yield (And as ratio of the reaction is 1:1), you will obtain 0.0673 moles of FeO.

Theoretical yield is given in grams, As molar mass of FeO is 71.844g/mol, theoretical yield of the reaction is:

0.0673 moles FeO × (71.844g / mol) =

<h3>4.84g of FeO is the theoretical yield</h3>
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