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To determine the coefficient of lithium fluoride (LiF) in the balanced chemical equation, we need to balance the chemical equation step by step. The unbalanced chemical equation is:
[tex]\[ \text{BF}_3 + \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
Step 1: Identify the number of each type of atom on both sides of the equation.
- On the reactant side:
- Boron (B): 1
- Fluorine (F): 3
- Lithium (Li): 2
- Sulfur (S): 1
- Oxygen (O): 3
- On the product side:
- Boron (B): 2
- Fluorine (F): 1
- Lithium (Li): 1
- Sulfur (S): 3
- Oxygen (O): 9
Step 2: Balance the boron atoms.
To balance boron, we need 2 boron atoms on each side. We can achieve this by placing a coefficient of 2 in front of [tex]\(\text{BF}_3\)[/tex]:
[tex]\[ 2\text{BF}_3 + \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
- Now, the updated counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on the reactant side, 1 on the product side
- Lithium (Li): 2 on the reactant side, 1 on the product side
- Sulfur (S): 1 on the reactant side, 3 on the product side
- Oxygen (O): 3 on the reactant side, 9 on the product side
Step 3: Balance the sulfur and oxygen atoms within [tex]\(\text{SO}_3\)[/tex].
We already have 3 sulfite groups ([tex]\(\text{SO}_3\)[/tex]) on the product side as part of [tex]\(\text{B}_2(\text{SO}_3)_3\)[/tex], which means we need 3 [tex]\(\text{Li}_2\text{SO}_3\)[/tex] on the reactant side. So, we place a coefficient of 3 in front of [tex]\(\text{Li}_2\text{SO}_3\)[/tex]:
[tex]\[ 2\text{BF}_3 + 3\text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
- Updated counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on the reactant side, 1 on the product side
- Lithium (Li): 6 on the reactant side, 1 on the product side
- Sulfur (S): 3 on both sides
- Oxygen (O): 9 on both sides
Step 4: Balance the fluorine and lithium atoms.
Since we have 6 fluorine atoms on the reactant side, we need 6 [tex]\(\text{LiF}\)[/tex] on the product side. Place a coefficient of 6 in front of [tex]\(\text{LiF}\)[/tex]:
[tex]\[ 2\text{BF}_3 + 3\text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6\text{LiF} \][/tex]
- Final counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on both sides
- Lithium (Li): 6 on both sides
- Sulfur (S): 3 on both sides
- Oxygen (O): 9 on both sides
The chemical equation is now balanced, and the coefficient of lithium fluoride ([tex]\(\text{LiF}\)[/tex]) is 6.
Therefore, the correct answer is [tex]\( \boxed{6} \)[/tex].
[tex]\[ \text{BF}_3 + \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
Step 1: Identify the number of each type of atom on both sides of the equation.
- On the reactant side:
- Boron (B): 1
- Fluorine (F): 3
- Lithium (Li): 2
- Sulfur (S): 1
- Oxygen (O): 3
- On the product side:
- Boron (B): 2
- Fluorine (F): 1
- Lithium (Li): 1
- Sulfur (S): 3
- Oxygen (O): 9
Step 2: Balance the boron atoms.
To balance boron, we need 2 boron atoms on each side. We can achieve this by placing a coefficient of 2 in front of [tex]\(\text{BF}_3\)[/tex]:
[tex]\[ 2\text{BF}_3 + \text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
- Now, the updated counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on the reactant side, 1 on the product side
- Lithium (Li): 2 on the reactant side, 1 on the product side
- Sulfur (S): 1 on the reactant side, 3 on the product side
- Oxygen (O): 3 on the reactant side, 9 on the product side
Step 3: Balance the sulfur and oxygen atoms within [tex]\(\text{SO}_3\)[/tex].
We already have 3 sulfite groups ([tex]\(\text{SO}_3\)[/tex]) on the product side as part of [tex]\(\text{B}_2(\text{SO}_3)_3\)[/tex], which means we need 3 [tex]\(\text{Li}_2\text{SO}_3\)[/tex] on the reactant side. So, we place a coefficient of 3 in front of [tex]\(\text{Li}_2\text{SO}_3\)[/tex]:
[tex]\[ 2\text{BF}_3 + 3\text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + \text{LiF} \][/tex]
- Updated counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on the reactant side, 1 on the product side
- Lithium (Li): 6 on the reactant side, 1 on the product side
- Sulfur (S): 3 on both sides
- Oxygen (O): 9 on both sides
Step 4: Balance the fluorine and lithium atoms.
Since we have 6 fluorine atoms on the reactant side, we need 6 [tex]\(\text{LiF}\)[/tex] on the product side. Place a coefficient of 6 in front of [tex]\(\text{LiF}\)[/tex]:
[tex]\[ 2\text{BF}_3 + 3\text{Li}_2\text{SO}_3 \rightarrow \text{B}_2(\text{SO}_3)_3 + 6\text{LiF} \][/tex]
- Final counts:
- Boron (B): 2 on both sides
- Fluorine (F): 6 on both sides
- Lithium (Li): 6 on both sides
- Sulfur (S): 3 on both sides
- Oxygen (O): 9 on both sides
The chemical equation is now balanced, and the coefficient of lithium fluoride ([tex]\(\text{LiF}\)[/tex]) is 6.
Therefore, the correct answer is [tex]\( \boxed{6} \)[/tex].
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