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To determine the coefficient of lithium fluoride ([tex]\(LiF\)[/tex]) in the balanced chemical equation, we need to follow a systematic approach of balancing each element in the reaction.
The unbalanced equation is:
[tex]\[ BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]
We'll balance the elements one by one:
1. Balance Boron (B):
- On the left side, boron appears in [tex]\(BF_3\)[/tex].
- On the right side, boron appears in [tex]\(B_2(SO_3)_3\)[/tex].
- There are 2 boron atoms in [tex]\(B_2(SO_3)_3\)[/tex]. Therefore, we need 2 [tex]\(BF_3\)[/tex] molecules to balance boron.
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]
2. Balance Fluorine (F):
- On the left side, fluorine appears in [tex]\(BF_3\)[/tex]. Each [tex]\(BF_3\)[/tex] molecule has 3 fluorine atoms.
- We have 2 [tex]\(BF_3\)[/tex] molecules, thus 2 x 3 = 6 fluorine atoms on the left side.
- On the right side, fluorine appears in [tex]\(LiF\)[/tex].
- To balance the 6 fluorine atoms, we need 6 [tex]\(LiF\)[/tex] molecules.
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
3. Balance Lithium (Li):
- On the left side, lithium appears in [tex]\(Li_2SO_3\)[/tex].
- Each [tex]\(Li_2SO_3\)[/tex] molecule has 2 lithium atoms.
- On the right side, lithium appears in [tex]\(LiF\)[/tex].
- We need 6 lithium atoms to balance the equation, so we need 3 [tex]\(Li_2SO_3\)[/tex] molecules.
[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
4. Balance Sulfur (S) and Oxygen (O):
- Let's double-check the sulfur and oxygen balance.
- On the left side, sulfur appears in [tex]\(Li_2SO_3\)[/tex] and we have 3 [tex]\(Li_2SO_3\)[/tex] molecules. Thus, there are [tex]\(3 \times 1 = 3\)[/tex] sulfur atoms and [tex]\(3 \times 3 = 9\)[/tex] oxygen atoms.
- On the right side, sulfur appears in [tex]\(B_2(SO_3)_3\)[/tex]. Each [tex]\(B_2(SO_3)_3\)[/tex] molecule has 3 sulfur atoms and [tex]\(3 \times 3 = 9\)[/tex] oxygen atoms.
- Both sulfur and oxygen are balanced.
The balanced chemical equation is:
[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
Thus, the coefficient of lithium fluoride ([tex]\(LiF\)[/tex]) in the balanced chemical reaction is [tex]\( \boxed{6} \)[/tex].
The unbalanced equation is:
[tex]\[ BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]
We'll balance the elements one by one:
1. Balance Boron (B):
- On the left side, boron appears in [tex]\(BF_3\)[/tex].
- On the right side, boron appears in [tex]\(B_2(SO_3)_3\)[/tex].
- There are 2 boron atoms in [tex]\(B_2(SO_3)_3\)[/tex]. Therefore, we need 2 [tex]\(BF_3\)[/tex] molecules to balance boron.
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + LiF \][/tex]
2. Balance Fluorine (F):
- On the left side, fluorine appears in [tex]\(BF_3\)[/tex]. Each [tex]\(BF_3\)[/tex] molecule has 3 fluorine atoms.
- We have 2 [tex]\(BF_3\)[/tex] molecules, thus 2 x 3 = 6 fluorine atoms on the left side.
- On the right side, fluorine appears in [tex]\(LiF\)[/tex].
- To balance the 6 fluorine atoms, we need 6 [tex]\(LiF\)[/tex] molecules.
[tex]\[ 2 BF_3 + Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
3. Balance Lithium (Li):
- On the left side, lithium appears in [tex]\(Li_2SO_3\)[/tex].
- Each [tex]\(Li_2SO_3\)[/tex] molecule has 2 lithium atoms.
- On the right side, lithium appears in [tex]\(LiF\)[/tex].
- We need 6 lithium atoms to balance the equation, so we need 3 [tex]\(Li_2SO_3\)[/tex] molecules.
[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
4. Balance Sulfur (S) and Oxygen (O):
- Let's double-check the sulfur and oxygen balance.
- On the left side, sulfur appears in [tex]\(Li_2SO_3\)[/tex] and we have 3 [tex]\(Li_2SO_3\)[/tex] molecules. Thus, there are [tex]\(3 \times 1 = 3\)[/tex] sulfur atoms and [tex]\(3 \times 3 = 9\)[/tex] oxygen atoms.
- On the right side, sulfur appears in [tex]\(B_2(SO_3)_3\)[/tex]. Each [tex]\(B_2(SO_3)_3\)[/tex] molecule has 3 sulfur atoms and [tex]\(3 \times 3 = 9\)[/tex] oxygen atoms.
- Both sulfur and oxygen are balanced.
The balanced chemical equation is:
[tex]\[ 2 BF_3 + 3 Li_2SO_3 \rightarrow B_2(SO_3)_3 + 6 LiF \][/tex]
Thus, the coefficient of lithium fluoride ([tex]\(LiF\)[/tex]) in the balanced chemical reaction is [tex]\( \boxed{6} \)[/tex].
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