Starting with 195 g Li2O and 106 g H2O, decide which reactant is present in limiting quantities. Given: Li2O+H2O→2LiOH
1 answer:
Taking into account the stoicionetry reaction and the definition of limiting reagent, Li₂O is present in limiting quantities.
The balanced reaction is:
Li₂O + H₂O → 2 LiOH
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:
- Li₂O: 1 mole
- H₂O: 1 mole
- LiOH: 2 moles
The molar mass of each compound is:
- Li₂O: 29.88 g/mole
- H₂O: 18 g/mole
- LiOH: 23.95 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Li₂O: 1 mole× 29.88 g/mole= 29.88 g
- H₂O: 1 mole× 18 g/mole= 18 g
- LiOH: 2 moles× 23.95 g/mole= 47.9 g
On the other side, 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 18 grams of H₂O reacts with 29.88 grams of Li₂O, 106 grams of H₂O react with how much mass of Li₂O?
mass of Li₂O= 175.96 grams
But 175.96 grams of Li₂O are not available, 195 grams are available. Since you have less moles than you need to react with 106 grams of H₂O, Li₂O will be the limiting reagent.
Finally, Li₂O is present in limiting quantities.
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