Question

Determine the percent yield for the reaction between 6.92 g K and 4.28g of oxygen gas if 7.36 g of potassium oxide is produced.

Answer 1

Answer:

percentage yield = 88.25%

Explanation:

Firstly, write the chemical reaction and balance the equation.

Potassium react with oxygen to produce potassium oxide.

K + 02 → K2O

Balance the equation

4K + 02 → 2K2O

The limiting reactant is K so the yield of potassium oxide can be calculated using grams for potassium.

atomic mass of K = 39.1g/mol

grams for 4 mole of potassium =  4(39.1) = 156.4 g

grams for 2 moles of K2O = 2( 39.1 × 2 + 16) = 188.4 g

If 156.4 g of K produces 188.4 g of K2O

6.92 g of K will produce ? gram of K2O

cross multiply

grams of K2O = 6.92 × 188.4/156.4

grams of K2O = 1303.72/156.4

grams of K2O = 8.33585677749

grams of K2O = 8.34 g

percentage yield = actual yield/theoretical yield × 100

actual yield = 7.36 g

theoretical yield = 8.34 g

percentage yield = 7.36/8.34 × 100

percentage yield =         736/8.34

percentage yield = 88.2494004796%

percentage yield = 88.25%

Answer 2

Taking into account definition of percent yield, the percent yield for the reaction is 88.25%.

Reaction stoichiometry

In first place, the balanced reaction is:

4 K + O₂ →2 K₂O

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:

• K: 4 moles
• O₂: 1 mole
• K₂O: 2 moles

The molar mass of the compounds is:

• K: 39 g/mole
• O₂: 32 g/mole
• K₂O: 94 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• K: 4 moles ×39 g/mole= 156 grams
• O₂: 1 mole ×32 g/mole= 32 grams
• K₂O: 2 moles ×55 g/mole= 188 grams

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.

Limiting reagent in this case

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 32 grams of O₂ reacts with 156 grams of K, 4.28 grams of O₂ reacts with how much mass of K?

$mass of K=\frac{4.28 grams of O_{2} x 156 grams of K}{32grams of O_{2}}$

mass of K= 20.865 grams

But 20.865 grams of K are not available, 6.92 grams are available. Since you have less mass than you need to react with 4.28 grams of O₂, K will be the limiting reagent.

Mass of K₂O formed

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 156 grams of K form 188 grams of K₂O, 6.92 grams of K form how much mass of K₂O?

$mass of K_{2} O=\frac{6.92 grams of Kx188 grams of K_{2} O}{156 grams of K}$

mass of K₂O= 8.34 grams

Then, 8.34 grams of K₂O can be produced from 6.92 grams of K and 4.28 grams of O₂.

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%:

$percent yield=\frac{actual yield}{theoretical yield}x100$

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.

Percent yield for the reaction in this case

In this case, you know:

• actual yield= 7.36 grams
• theorical yield= 8.34 grams

Replacing in the definition of percent yields:

$percent yield=\frac{7.36 grams}{8.34 grams}x100$

Solving:

percent yield= 88.25%

Finally, the percent yield for the reaction is 88.25%.

Learn more about

the reaction stoichiometry:

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percent yield:

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