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Sagot :
Let's go through the given chemical equation step by step to determine if it is balanced:
[tex]\[ \text{ZnSO}_4 + \text{Li}_2\text{CO}_3 \rightarrow \text{ZnCO}_3 + \text{Li}_2\text{SO}_4 \][/tex]
First, let's count the number of atoms of each element on the reactant and product sides.
### Reactant Side:
1. Zn (Zinc):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] has 1 Zn atom.
2. S (Sulfur):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] has 1 S atom.
3. O (Oxygen):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] contains 4 O atoms.
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] contains 3 O atoms.
- So, the total number of O atoms on the reactant side is [tex]\( 4 + 3 = 7 \)[/tex] atoms.
4. Li (Lithium):
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] has 2 Li atoms.
5. C (Carbon):
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] has 1 C atom.
### Product Side:
1. Zn (Zinc):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] has 1 Zn atom.
2. S (Sulfur):
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] has 1 S atom.
3. O (Oxygen):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] contains 3 O atoms.
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] contains 4 O atoms.
- So, the total number of O atoms on the product side is [tex]\( 3 + 4 = 7 \)[/tex] atoms.
4. Li (Lithium):
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] has 2 Li atoms.
5. C (Carbon):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] has 1 C atom.
When we compare the counts:
- Zn: 1 atom on both sides.
- S: 1 atom on both sides.
- O: 7 atoms on both sides.
- Li: 2 atoms on both sides.
- C: 1 atom on both sides.
Since the counts for each type of atom are the same on both sides of the equation, the equation is already balanced.
Therefore, the statement that best describes the balanced state of the equation is:
- Atoms in the equation are already in balance.
[tex]\[ \text{ZnSO}_4 + \text{Li}_2\text{CO}_3 \rightarrow \text{ZnCO}_3 + \text{Li}_2\text{SO}_4 \][/tex]
First, let's count the number of atoms of each element on the reactant and product sides.
### Reactant Side:
1. Zn (Zinc):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] has 1 Zn atom.
2. S (Sulfur):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] has 1 S atom.
3. O (Oxygen):
- The reactant [tex]\( \text{ZnSO}_4 \)[/tex] contains 4 O atoms.
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] contains 3 O atoms.
- So, the total number of O atoms on the reactant side is [tex]\( 4 + 3 = 7 \)[/tex] atoms.
4. Li (Lithium):
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] has 2 Li atoms.
5. C (Carbon):
- The reactant [tex]\( \text{Li}_2\text{CO}_3 \)[/tex] has 1 C atom.
### Product Side:
1. Zn (Zinc):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] has 1 Zn atom.
2. S (Sulfur):
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] has 1 S atom.
3. O (Oxygen):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] contains 3 O atoms.
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] contains 4 O atoms.
- So, the total number of O atoms on the product side is [tex]\( 3 + 4 = 7 \)[/tex] atoms.
4. Li (Lithium):
- The product [tex]\( \text{Li}_2\text{SO}_4 \)[/tex] has 2 Li atoms.
5. C (Carbon):
- The product [tex]\( \text{ZnCO}_3 \)[/tex] has 1 C atom.
When we compare the counts:
- Zn: 1 atom on both sides.
- S: 1 atom on both sides.
- O: 7 atoms on both sides.
- Li: 2 atoms on both sides.
- C: 1 atom on both sides.
Since the counts for each type of atom are the same on both sides of the equation, the equation is already balanced.
Therefore, the statement that best describes the balanced state of the equation is:
- Atoms in the equation are already in balance.
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