How is the equilibrium constant of a reaction determined?
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Taking into account definition of percent yield, the percent yield for the reaction is 84.88%.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al: 2 moles
- H₂SO₄: 3 moles
- Al₂(SO₄)₃. 1 mole
- H₂: 3 moles
The molar mass of the compounds is:
- Al: 27 g/mole
- H₂SO₄: 98 g/mole
- Al₂(SO₄)₃: 342 g/mole
- H₂: 2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al: 2 moles ×27 g/mole= 54 grams
- H₂SO₄: 3 moles ×98 g/mole= 294 grams
- Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
- H₂: 3 moles ×2 g/mole= 6 grams
The following rule of three can be applied: if by reaction stoichiometry 54 grams of aluminium form 342 grams of aluminium sulfate, 14 grams of aluminium form how much mass of aluminium sulfate?
<u><em>mass of aluminium sulfate= 88.67 grams</em></u>
Then, 88.67 grams of aluminium sulfate can be produced if 14.0 g of aluminium reacts with excess sulfuric acid.
<h3>Percent yield</h3>The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
<h3>Percent yield for the reaction in this case</h3>In this case, you know:
- actual yield= 75.26 grams
- theorical yield= 88.67 grams
Replacing in the definition of percent yields:
Solving:
<u><em>percent yield= 84.88%</em></u>
Finally, the percent yield for the reaction is 84.88%.
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