Answer:
45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂
Step-by-step explanation:
The balanced reaction is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- H₂: 2 moles
- O₂: 1 mole
- H₂O: 2 moles
Being the molar mass of the compounds:
- H₂: 2 g/mole
- O₂: 32 g/mole
- H₂O: 18 g/mole
then, by reaction stoichiometry, the following amounts of reactant and product mass participate:
- H₂: 2 moles* 2 g/mole= 4 g
- O₂: 1 mole* 32 g/mole= 32 g
- H₂O: 2 moles* 18 g/mole= 36 g
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 reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 4 g of H₂ react with 32 g of O₂, 36.8 g of H₂ with how much mass of O₂ will it react?
mass of O₂=294.4 grams
But 294.4 grams of O₂ are not available, 40.2 grams are available. Since you have less mass than you need to react with 36.8 grams of H₂, oxygen O₂ will be the limiting reagent.
Then you can apply the following rule of three: if by stoichiometry 32 grams of O₂ form 36 grams of H₂O, 40.2 grams of O₂ how much mass of H₂O will it form?
mass of H₂O= 45.225 grams
45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂