Question

Starting with 77.9 g Li₂O and 76.5 g H₂O, decide which reactant is present in limiting quantities in the reaction:

Li₂O + H₂O → 2LiOH

a) Li₂O
b) H₂O
c) Insufficient data
d) None of the above

Answer 1

Upon comparing the moles of lithium oxide (Li₂O) and water (H₂O) to the balanced equation for their reaction, lithium oxide is identified as the limiting reactant because it has fewer moles than required to react with the available moles of water.

Step-by-step explanation:

To determine the limiting reactant in the chemical reaction between lithium oxide (Li₂O) and water (H₂O), one must convert the mass of each reactant to moles using their respective molar masses. According to the reaction stoichiometry:

1. Calculate the moles of Li₂O using its molar mass (29.88 g/mol).
2. Calculate the moles of H₂O using its molar mass (18.02 g/mol).
3. Compare the mole ratio obtained from the calculation to the stoichiometric ratio required by the balanced chemical equation.

The molar mass of lithium oxide (Li₂O) is 29.88 g/mol. By dividing 77.9 g Li₂O by 29.88 g/mol, we obtain approximately 2.61 moles of Li₂O.

The molar mass of water (H₂O) is 18.02 g/mol. By dividing 76.5 g H₂O by 18.02 g/mol, we obtain approximately 4.25 moles of H₂O. Since the stoichiometry of the reaction shows 1 mole of Li₂O reacts with 1 mole of H₂O to produce 2 moles of LiOH, we can conclude that the moles of Li₂O are less than the moles of H₂O.

Therefore, Li₂O is the limiting reactant because it will be consumed first, according to the stoichiometric proportions of the reactants in the balanced equation.