Get clear, concise, and accurate answers to your questions on IDNLearn.com. Ask anything and receive thorough, reliable answers from our community of experienced professionals.
Sagot :
To determine which chemical equation obeys the law of conservation of mass, we need to ensure that the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Let's analyze each given equation:
### Option A: \(2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 7 CO_2 + CCl_4 + 10 H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(12 \times 2 = 24\)
Products:
- C: \(7 + 1 = 8\)
- H: \(10 \times 2 = 20\)
- Cl: \(4\)
- O: \(7 \times 2 + 10 = 14 + 10 = 24\)
Balanced: Yes
### Option B: \(C_4H_{10} + Cl_2 + 2 O_2 \rightarrow 7 CO_2 + CCl_4 + 8 H_2O\)
Reactants:
- C: \(4\)
- H: \(10\)
- Cl: \(2\)
- O: \(2 \times 2 = 4\)
Products:
- C: \(7 + 1 = 8\)
- H: \(8 \times 2 = 16\)
- Cl: \(4\)
- O: \(7 \times 2 + 8 = 14 + 8 = 22\)
Balanced: No
### Option C: \(2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 4 CO_2 + CCl_4 + H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(12 \times 2 = 24\)
Products:
- C: \(4 + 1 = 5\)
- H: \(1 \times 2 = 2\)
- Cl: \(4\)
- O: \(4 \times 2 + 1 = 8 + 1 = 9\)
Balanced: No
### Option D: \(2 C_4H_{10} + 2 Cl_2 + 6 O_2 \rightarrow CO_2 + CCl_4 + 6 H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(6 \times 2 = 12\)
Products:
- C: \(1 + 1 = 2\)
- H: \(6 \times 2 = 12\)
- Cl: \(4\)
- O: \(1 \times 2 + 6 = 2 + 6 = 8\)
Balanced: No
### Option E: \(4 C_4H_{10} + 4 Cl_2 + 14 O_2 \rightarrow 14 CO_2 + CCl_4 + 20 H_2O\)
Reactants:
- C: \(4 \times 4 = 16\)
- H: \(4 \times 10 = 40\)
- Cl: \(4 \times 2 = 8\)
- O: \(14 \times 2 = 28\)
Products:
- C: \(14 + 1 = 15\)
- H: \(20 \times 2 = 40\)
- Cl: \(4\)
- O: \(14 \times 2 + 20 = 28 + 20 = 48\)
Balanced: No
### Conclusion
The reaction that follows the law of conservation of mass is:
[tex]\[ 2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 7 CO_2 + CCl_4 + 10 H_2O \][/tex]
So, the correct answer is Option A.
### Option A: \(2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 7 CO_2 + CCl_4 + 10 H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(12 \times 2 = 24\)
Products:
- C: \(7 + 1 = 8\)
- H: \(10 \times 2 = 20\)
- Cl: \(4\)
- O: \(7 \times 2 + 10 = 14 + 10 = 24\)
Balanced: Yes
### Option B: \(C_4H_{10} + Cl_2 + 2 O_2 \rightarrow 7 CO_2 + CCl_4 + 8 H_2O\)
Reactants:
- C: \(4\)
- H: \(10\)
- Cl: \(2\)
- O: \(2 \times 2 = 4\)
Products:
- C: \(7 + 1 = 8\)
- H: \(8 \times 2 = 16\)
- Cl: \(4\)
- O: \(7 \times 2 + 8 = 14 + 8 = 22\)
Balanced: No
### Option C: \(2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 4 CO_2 + CCl_4 + H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(12 \times 2 = 24\)
Products:
- C: \(4 + 1 = 5\)
- H: \(1 \times 2 = 2\)
- Cl: \(4\)
- O: \(4 \times 2 + 1 = 8 + 1 = 9\)
Balanced: No
### Option D: \(2 C_4H_{10} + 2 Cl_2 + 6 O_2 \rightarrow CO_2 + CCl_4 + 6 H_2O\)
Reactants:
- C: \(2 \times 4 = 8\)
- H: \(2 \times 10 = 20\)
- Cl: \(2 \times 2 = 4\)
- O: \(6 \times 2 = 12\)
Products:
- C: \(1 + 1 = 2\)
- H: \(6 \times 2 = 12\)
- Cl: \(4\)
- O: \(1 \times 2 + 6 = 2 + 6 = 8\)
Balanced: No
### Option E: \(4 C_4H_{10} + 4 Cl_2 + 14 O_2 \rightarrow 14 CO_2 + CCl_4 + 20 H_2O\)
Reactants:
- C: \(4 \times 4 = 16\)
- H: \(4 \times 10 = 40\)
- Cl: \(4 \times 2 = 8\)
- O: \(14 \times 2 = 28\)
Products:
- C: \(14 + 1 = 15\)
- H: \(20 \times 2 = 40\)
- Cl: \(4\)
- O: \(14 \times 2 + 20 = 28 + 20 = 48\)
Balanced: No
### Conclusion
The reaction that follows the law of conservation of mass is:
[tex]\[ 2 C_4H_{10} + 2 Cl_2 + 12 O_2 \rightarrow 7 CO_2 + CCl_4 + 10 H_2O \][/tex]
So, the correct answer is Option A.
Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.