Taking into account the reaction stoichiometry, 2.4041 grams of K₂O can be produced from 2 grams of K and 5 grams of O₂.
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: 39.1 g/mole
- O₂: 32 g/mole
- K₂O: 94.2 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- K: 4 moles ×39.1 g/mole= 156.4 grams
- O₂: 1 mole ×32 g/mole= 32 grams
- K₂O: 2 moles ×94.2 g/mole= 188.4 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 156.4 grams of K, 5 grams of O₂ reacts with how much mass of K?
mass of K= (156.4 grams of K× 5 grams of O₂)÷ 32 grams of O₂
mass of K= 24.4375 grams
But 24.4375 grams of K are not available, 2 grams are available. Since you have less mass than you need to react with 5 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 156.4 grams of K form 188 grams of K₂O, 2 grams of K form how much mass of K₂O?
mass of K₂O= (2 grams of K× 188 grams of K₂O)÷ 156.4 grams of K
mass of K₂O= 2.4041 grams
Then, 2.4041 grams of K₂O can be produced from 2 grams of K and 5 grams of O₂.
Learn more about the reaction stoichiometry:
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