In the reaction: Li + F_2 → LiF, what is the coefficient of lithium fluoride (LiF) in the balanced chemical equation?

• A. 1
• B. 3
• C. 4
• D. 2

Answer: (D)

Verified step by step guidance

1. Write the unbalanced chemical equation: \(\mathrm{Li} + \mathrm{F}_2 \rightarrow \mathrm{LiF}\).
2. Identify the number of atoms of each element on both sides. On the left, fluorine is diatomic (\(\mathrm{F}_2\)), so there are 2 fluorine atoms, while lithium is a single atom. On the right, lithium fluoride (\(\mathrm{LiF}\)) contains 1 lithium and 1 fluorine atom per formula unit.
3. Balance fluorine atoms first by placing a coefficient of 2 in front of \(\mathrm{LiF}\) to have 2 fluorine atoms on the product side: \(\mathrm{Li} + \mathrm{F}_2 \rightarrow 2\mathrm{LiF}\).
4. Balance lithium atoms by placing a coefficient of 2 in front of \(\mathrm{Li}\) on the reactant side to match the 2 lithium atoms in the products: \$2\mathrm{Li} + \mathrm{F}_2 \rightarrow 2\mathrm{LiF}$.
5. Verify that the number of atoms for each element is equal on both sides, confirming the equation is balanced with a coefficient of 2 for lithium fluoride.