Consider a 5.430-g mixture of FeO and Fe3O

[textbooks]

Consider a 5.430-g mixture of FeO and Fe3O4. You react this mixture with excess of oxygen to form 5.779g Fe2O3. Calculate the percent by mass of FeO in the original mixture.

 

[cdotter]

You can start by writing a balanced chemical equation for the reaction.

 

[sjb-2812]

You can start by writing a balanced chemical equation for the reaction.

Writing one for the reactions, might be better :)

 

[Borek]

Writing TWO for reactions will be even better.

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[LudusRex]

To calculate the percent by mass of FeO in the original mixture, we first need to determine the moles of Fe2O3 formed from the reaction. From the given information, we know that the total mass of the mixture before the reaction was 5.430 g. After the reaction, the mass of the mixture increased to 5.779 g, indicating a mass difference of 0.349 g. This mass difference is equal to the mass of oxygen that reacted with the mixture.

Next, we need to determine the moles of oxygen that reacted. The molar mass of oxygen is 16.00 g/mol, so the number of moles of oxygen is 0.349 g / 16.00 g/mol = 0.0218 mol.

Since oxygen is in excess, we can assume that all of the oxygen reacted with the FeO and Fe3O4 to form Fe2O3. Therefore, the moles of Fe2O3 formed is also equal to 0.0218 mol.

Now, we can use the balanced chemical equation for the reaction to determine the moles of FeO present in the original mixture:

FeO + 1/2 O2 → Fe2O3

From the equation, we can see that 1 mole of FeO reacts with 1/2 mole of O2 to form 1 mole of Fe2O3. Therefore, the moles of FeO in the original mixture is equal to 0.0218 mol.

To calculate the percent by mass of FeO, we need to compare the mass of FeO to the total mass of the original mixture:

% by mass of FeO = (0.0218 mol FeO) x (71.85 g/mol FeO) / (5.430 g mixture) x 100%

= 0.312% FeO

Therefore, the percent by mass of FeO in the original mixture is 0.312%.