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Sagot :
To determine which of the given chemical equations is balanced, we need to check the stoichiometry of each element in the reactants and products for each equation. A balanced equation has the same number of each type of atom on both sides. Let’s go through each option one-by-one:
### Option A: [tex]\( \mathrm{NaClO_3 \rightarrow NaCl + 3 O_2} \)[/tex]
1. Sodium (Na):
- Reactant: 1 Na
- Product: 1 Na
- Balanced for Na.
2. Chlorine (Cl):
- Reactant: 1 Cl
- Product: 1 Cl
- Balanced for Cl.
3. Oxygen (O):
- Reactant: 3 O
- Product: 6 O (because [tex]\( 3 \times O_2 \)[/tex] means 3 molecules of [tex]\( O_2 \)[/tex], hence [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms)
- Unbalanced for O (Reactant: 3 O, Product: 6 O).
Hence, Option A is unbalanced.
### Option B: [tex]\( \mathrm{2 HgO \rightarrow 2 Hg + O_2} \)[/tex]
1. Mercury (Hg):
- Reactant: 2 Hg
- Product: 2 Hg
- Balanced for Hg.
2. Oxygen (O):
- Reactant: 2 O
- Product: 2 O (because [tex]\( O_2 \)[/tex] means 2 oxygen atoms)
- Balanced for O.
Hence, Option B is balanced.
### Option C: [tex]\( \mathrm{2 ZnS + O_2 \rightarrow 2 ZnO + 2 SO_2} \)[/tex]
1. Zinc (Zn):
- Reactant: 2 Zn
- Product: 2 Zn
- Balanced for Zn.
2. Sulfur (S):
- Reactant: 2 S
- Product: 2 S
- Balanced for S.
3. Oxygen (O):
- Reactant: 2 O
- Product: 6 O (because [tex]\( 2 \times (ZnO) \)[/tex] and [tex]\( 2 \times (SO_2) \)[/tex], where each contains 1 and 2 oxygens respectively. Thus [tex]\( 2 \times 1 + 2 \times 2 = 2 + 4 = 6 \)[/tex]).
- Unbalanced for O (Reactant: 2 O, Product: 6 O).
Hence, Option C is unbalanced.
### Option D: [tex]\( \mathrm{4 CO + Fe_3O_4 \rightarrow 4 CO_2 + Fe} \)[/tex]
1. Carbon (C):
- Reactant: 4 C (from [tex]\( 4 \times \text{CO} \)[/tex])
- Product: 4 C (from [tex]\( 4 \times \text{CO}_2 \)[/tex])
- Balanced for C.
2. Oxygen (O):
- Reactant: 4 O (from [tex]\( 4 \times \text{CO} \)[/tex]) + 4 O (from [tex]\( \text{Fe}_3\text{O}_4 \)[/tex]) = 8 O
- Product: 8 O (from [tex]\( 4 \times \text{CO}_2 \)[/tex])
- Balanced for O.
3. Iron (Fe):
- Reactant: 3 Fe
- Product: 1 Fe
- Unbalanced for Fe.
Hence, Option D is unbalanced.
### Option E: [tex]\( \mathrm{2 Fe + 2 Cl_2 \rightarrow 2 FeCl_3} \)[/tex]
1. Iron (Fe):
- Reactant: 2 Fe
- Product: 2 Fe
- Balanced for Fe.
2. Chlorine (Cl):
- Reactant: 4 Cl (from [tex]\( 2 \times \text{Cl}_2 \)[/tex])
- Product: 6 Cl (from [tex]\( 2 \times \text{FeCl}_3 \)[/tex])
- Unbalanced for Cl (Reactant: 4 Cl, Product: 6 Cl).
Hence, Option E is unbalanced.
### Conclusion
The correctly balanced equation among the options given is:
Option B: [tex]\( \mathrm{2 HgO \rightarrow 2 Hg + O_2} \)[/tex].
### Option A: [tex]\( \mathrm{NaClO_3 \rightarrow NaCl + 3 O_2} \)[/tex]
1. Sodium (Na):
- Reactant: 1 Na
- Product: 1 Na
- Balanced for Na.
2. Chlorine (Cl):
- Reactant: 1 Cl
- Product: 1 Cl
- Balanced for Cl.
3. Oxygen (O):
- Reactant: 3 O
- Product: 6 O (because [tex]\( 3 \times O_2 \)[/tex] means 3 molecules of [tex]\( O_2 \)[/tex], hence [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms)
- Unbalanced for O (Reactant: 3 O, Product: 6 O).
Hence, Option A is unbalanced.
### Option B: [tex]\( \mathrm{2 HgO \rightarrow 2 Hg + O_2} \)[/tex]
1. Mercury (Hg):
- Reactant: 2 Hg
- Product: 2 Hg
- Balanced for Hg.
2. Oxygen (O):
- Reactant: 2 O
- Product: 2 O (because [tex]\( O_2 \)[/tex] means 2 oxygen atoms)
- Balanced for O.
Hence, Option B is balanced.
### Option C: [tex]\( \mathrm{2 ZnS + O_2 \rightarrow 2 ZnO + 2 SO_2} \)[/tex]
1. Zinc (Zn):
- Reactant: 2 Zn
- Product: 2 Zn
- Balanced for Zn.
2. Sulfur (S):
- Reactant: 2 S
- Product: 2 S
- Balanced for S.
3. Oxygen (O):
- Reactant: 2 O
- Product: 6 O (because [tex]\( 2 \times (ZnO) \)[/tex] and [tex]\( 2 \times (SO_2) \)[/tex], where each contains 1 and 2 oxygens respectively. Thus [tex]\( 2 \times 1 + 2 \times 2 = 2 + 4 = 6 \)[/tex]).
- Unbalanced for O (Reactant: 2 O, Product: 6 O).
Hence, Option C is unbalanced.
### Option D: [tex]\( \mathrm{4 CO + Fe_3O_4 \rightarrow 4 CO_2 + Fe} \)[/tex]
1. Carbon (C):
- Reactant: 4 C (from [tex]\( 4 \times \text{CO} \)[/tex])
- Product: 4 C (from [tex]\( 4 \times \text{CO}_2 \)[/tex])
- Balanced for C.
2. Oxygen (O):
- Reactant: 4 O (from [tex]\( 4 \times \text{CO} \)[/tex]) + 4 O (from [tex]\( \text{Fe}_3\text{O}_4 \)[/tex]) = 8 O
- Product: 8 O (from [tex]\( 4 \times \text{CO}_2 \)[/tex])
- Balanced for O.
3. Iron (Fe):
- Reactant: 3 Fe
- Product: 1 Fe
- Unbalanced for Fe.
Hence, Option D is unbalanced.
### Option E: [tex]\( \mathrm{2 Fe + 2 Cl_2 \rightarrow 2 FeCl_3} \)[/tex]
1. Iron (Fe):
- Reactant: 2 Fe
- Product: 2 Fe
- Balanced for Fe.
2. Chlorine (Cl):
- Reactant: 4 Cl (from [tex]\( 2 \times \text{Cl}_2 \)[/tex])
- Product: 6 Cl (from [tex]\( 2 \times \text{FeCl}_3 \)[/tex])
- Unbalanced for Cl (Reactant: 4 Cl, Product: 6 Cl).
Hence, Option E is unbalanced.
### Conclusion
The correctly balanced equation among the options given is:
Option B: [tex]\( \mathrm{2 HgO \rightarrow 2 Hg + O_2} \)[/tex].
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