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An unbalanced chemical equation for the reaction of boron fluoride with lithium sulfite is shown below:

[tex]\[ BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]

What is the coefficient of lithium fluoride in the balanced chemical reaction?

A. 1
B. 3
C. 4
D. 6

Sagot :

GinnyAnswer
To balance the provided chemical equation:

[tex]\[ BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]

we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's break it down step-by-step.

1. Identify the number of atoms of each element in the reactants and products:
- Reactants:
- [tex]\(BF_3\)[/tex]: 1 B, 3 F
- [tex]\(Li_2SO_3\)[/tex]: 2 Li, 1 S, 3 O
- Products:
- [tex]\(B_2(SO_3)_3\)[/tex]: 2 B, 3 S, 9 O
- [tex]\(LiF\)[/tex]: 1 Li, 1 F

2. Balance boron (B) atoms:
We have 2 B atoms in [tex]\(B_2(SO_3)_3\)[/tex], so we need 2 [tex]\(BF_3\)[/tex] molecules on the reactants side:
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]
- Boron is now balanced: 2 B atoms on both sides.

3. Balance fluorine (F) atoms:
With 2 [tex]\(BF_3\)[/tex], we have [tex]\(2 \times 3 = 6\)[/tex] F atoms on the reactants side. We need 6 F atoms on the products side:
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
- Fluorine is now balanced: 6 F atoms on both sides.

4. Balance lithium (Li) atoms:
On the products side, we now have 6 Li atoms from 6 [tex]\(LiF\)[/tex]. Thus, we need 3 [tex]\(Li_2SO_3\)[/tex] molecules on the reactants side to provide the 6 lithium atoms:
[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
- Lithium is now balanced: 6 Li atoms on both sides.

5. Balance sulfur (S) atoms:
We have 3 sulfur atoms from [tex]\(B_2(SO_3)_3\)[/tex] on the products side and 3 [tex]\(Li_2SO_3\)[/tex] on the reactants side, which provides 3 sulfur atoms:
- Sulfur is now balanced: 3 S atoms on both sides.

6. Balance oxygen (O) atoms:
The oxygen atoms are now automatically balanced as:
- Reactants: 3 [tex]\(Li_2SO_3\)[/tex] each contribute 3 oxygen atoms, leading to [tex]\(3 \times 3 = 9\)[/tex] oxygen atoms.
- Products: [tex]\(B_2(SO_3)_3\)[/tex] has 9 oxygen atoms.

The balanced equation is:

[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]

The coefficient of lithium fluoride ([tex]\(LiF\)[/tex]) in the balanced chemical reaction is 6.
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