Question

If 73.5 grams of FeS2 is allowed to react with 44 grams of O2 according to the following equation, how many grams of Fe2O3 are produced? FeS2 + O2 → Fe2O3 + SO2

Answer 1

Balanced equation:

4 FeS₂ + 11 O₂ --> 2 Fe₂O₃ + 8 SO₂

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Find the limiting reactant (the one that produces the least amount of product) :

(73.5 g FeS₂) * (1 mole FeS₂ / 119.975 g FeS₂) * (2 moles Fe₂O₃ / 4 moles FeS2) * (159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)

= 48.9 g Fe₂O₃ …………. to 3 sig figs

(44 g O₂) * (1 mole O₂ / 31.9988 g O₂) * (2 moles Fe₂O₃ / 11 moles O₂) *  

(159.6882 g Fe₂O₃ / 1 mole Fe₂O₃)

= 40 g Fe₂O₃ ………………. to 2 sig figs

O₂ is the limiting reactant and the theoretical yield is 44 g Fe₂O₃

Answer 2

Answer:

39.925 grams of Fe2O3 are produced.

Explanation:

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

To comply with this law, the equation must be balanced. So, in this case, the balanced equation is:

4 FeS₂ + 11 O₂ --> 2 Fe₂O₃ + 8 SO₂

  First of all it is convenient to determine the molar mass of each compound knowing the atomic mass of each element:

• Fe: 55.85 g/mol
• S: 32 g/mol
• O: 16 g/mol

Then:

• FeS₂: 55.85 g/mol + 2*32 g/mol= 119.85 g/mol
• O₂: 2* 16 g/mol= 32 g/mol
• Fe₂O₃: 2* 55.85 g/mol + 3*16 g/mol= 159.7 g/mol
• SO₂: 32 g/mol+ 2*16 g/mol=  64 g/mol

Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you can see the amount in moles of reagents and products that participate in the reaction:

• FeS₂: 4 moles
• O₂: 11 moles
• Fe₂O₃: 2 moles
• SO₂: 8 moles

Knowing the molar mass of each reagent and product it is possible to determine the mass that react and are formed in the reaction:

• FeS₂: 119.85 g/mol *4 moles= 479.4 g
• O₂: 32 g/mol *11 moles= 352 g
• Fe₂O₃: 159.7 g/mol *2 moles= 319.4  g
• SO₂: 64 g/mol *8 moles= 512 g

The limiting reagent is now calculated.

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the stoichiometry of the reaction and a simple rule of three as follows:

if by stoichiometry 479.4 g of FeS₂ reacts with 352 g of O₂, if 73.5 g of FeS₂ react how many grams of O₂ will be necessary?

$mass of O_{2} =\frac{73.5 g FeS_{2} *352gO_{2} }{479.4 g FeS_{2}}$

mass of O₂≅ 53.96 g

But 53.96 grams of O₂ are not available, 44 grams are available. Since you have less moles than you need to react with 73.5 grams of FeS₂, oxygen O₂ will be the limiting reagent.

Then, as the reaction will stop when the limiting reagent is over, this reagent will be useful for subsequent calculations.

To determine the amount of Fe₂O₃ produced, the following rule of three applies: if by stoichiometry 352 grams of O₂ produce 319.4 grams of Fe₂O₃, 44 grams of O₂ how many grams of Fe₂O₃ will they produce?

$mass of Fe_{2} O_{3} =\frac{44gO_{2}*319.4gFe_{2} O_{3} }{352gO_{2} }$

mass of Fe₂O₃=39.925 grams

39.925 grams of Fe2O3 are produced.