Question

For the reaction 2Li(s)+F₂(g) → 2LiF(s) identify the limiting reactant for each of the initial quantities of reactants. Drag the appropriate items to their respective bins.

Items:
2.85X10³gLi
6.79X10³ gF₂
10.5gLi
37.2gF₂
1.0gLi
1.0gF₂

Li is the limiting reagent F₂ is the limiting reagent

Answer 1

To find the limiting reactant for the reaction 2Li(s) + F₂(g) → 2LiF(s), calculate the moles of Li and F₂ from the given masses and compare to the stoichiometric mole ratio; the reactant in deficit is the limiting one.

Step-by-step explanation:

To identify the limiting reactant for the chemical reaction 2Li(s) + F₂(g) → 2LiF(s), we must perform calculations for each set of initial quantities given:

1. Calculate the number of moles of Li and F₂ using their respective molar masses (Li = 6.94 g/mol, F₂ = 38 g/mol).
2. Using the stoichiometry of the balanced equation, determine the mole ratio needed for the reaction (2 moles of Li per 1 mole of F₂).
3. Compare the actual mole ratios from the given masses to the stoichiometric ratio to identify the limiting reactant.

The limiting reactant is the one that will run out first based on these ratios, thus determining the amount of product that can be formed.