Taking into account definition of reaction stoichiometry and percent yield, the percent yield for the reaction is 11.72%.
Reaction stoichiometry
In first place, the balanced reaction is:
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
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:
- H₂SO₄: 1 mole
- KOH: 2 moles
- K₂SO₄: 1 mole
- H₂O: 2 moles
The molar mass of the compounds is:
- H₂SO₄: 98 g/mole
- KOH: 56.1 g/mole
- K₂SO₄: 174.2 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- H₂SO₄: 1 mole ×98 g/mole= 98 grams
- KOH: 2 moles ×56.1 g/mole= 112.2 grams
- K₂SO₄: 1 mole ×174.2 g/mole= 174.2 grams
- H₂O: 2 moles ×18 g/mole= 36 grams
Rule of three
The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied.
To solve a direct rule of three, the following formula must be followed, being a, b and c known data and x the variable to be calculated:
a ⇒ b
c ⇒ x
So: x=(c×b)÷a
Limiting reagent
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 a simple rule of three as follows: if by stoichiometry 98 grams of H₂SO₄ reacts with 112.2 grams of KOH, 11 grams of H₂SO₄ reacts with how much mass of KOH?
mass of KOH= (11 grams of H₂SO₄×112.2 grams of KOH,)÷98 grams of H₂SO₄
mass of KOH= 12.59 grams
But 12.59 grams of KOH are not available, 11 grams are available. Since you have less mass than you need to react with 11 grams of H₂SO₄, KOH will be the limiting reagent.
Mass of K₂SO₄ formed
Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 112.2 grams of KOH form 174.2 grams of K₂SO₄, 11 grams of KOH form how much mass of K₂SO₄?
mass of K₂SO₄= (11 grams of KOH× 174.2 grams of K₂SO₄)÷ 112.2 grams of KOH
mass of K₂SO₄= 17.07 grams
Percent yield
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%:
percent yield= (actual yield÷ theoretical yield)×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.
In this case, you know:
actual yield= 2 grams
theorical yield= 17.07 grams
Replacing in the definition of percent yields:
percent yield= (2 grams÷ 17.07 grams)×100%
Solving:
percent yield= 11.72%
Finally, the percent yield is 11.72%.