Chemistry, 18.10.2020 09:01 maya7544

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

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Answer from: rwbrayan8727

Explanation:

A 5.430 g mixture of FeO and Fe3O4 is reacted with excess of oxygen to form 5.779 g Fe2O3. Find the masses of FeO and Fe3O4 present in the mixture.

I found the answer by doing the following:

Fe2O3 = 159.69 g/mol

FeO = 71.745 g/mol

Fe3O4 = 231.535 g/mol

5.779 g Fe2O3 / 159.69 g/mol = 0.03619 mol Fe2O3

x + y :rarrow: 0.03619 mol Fe2O3, where x = FeO and y = Fe3O4.

This implies that y = 0.03619 - x.

5.430 g = x*71.745 g/mol + (0.03619 - x)*231.535g/mol

x = 0.01846 mol * 71.745 g/mol = 1.324 g FeO

y = 0.03619 - 0.01846 = 0.01773 mol * 231.535 g/mol = 4.105 g Fe3O4

My question is why doesn't the following work:

4FeO + O2 :rarrow: 2Fe2O3

4Fe3O4 + O2 :rarrow: 6Fe2O3

2FeO + 2Fe3O4 + O2 :rarrow: 4Fe2O3

5.779 g Fe2O3 / 159.69 g/mol = 0.03619 mol Fe2O3

Product:reactants in question are in a 2:1 ratio as given by the stoichiometric coefficients.

0.03619 mol / 2 = 0.01810 mol of FeO and 0.01810 mol Fe3O4.

0.01810 mol FeO * 71.745 g/mol = 1.299 g FeO

0.01810 mol Fe3O4 * 231.535 g/mol = 4.191 g Fe3O4

Obviously this doesn't work because the original mixture is 5.430 g and not 1.299 g + 4.191 g = 5.490 g. Why doesn't this work the way I think it should? Thank you.

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

Answer from: Quest

gas and precipitate r the correct answers

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Consider a 5.430 g mixture of FeO and Fe3O4. You react this mixture with excess of oxygen to form 5....

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