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To balance the chemical equation [tex]\(C_2H_6 + O_2 \rightarrow CO_2 + H_2O\)[/tex], we need to ensure that the number of each type of atom on the reactant side (left side) is equal to the number of each type of atom on the product side (right side). We'll do this step-by-step:
### Step 1: Count the Atoms
1. Reactants:
- [tex]\(C_2H_6\)[/tex]: 2 Carbon (C) atoms, 6 Hydrogen (H) atoms
- [tex]\(O_2\)[/tex]: 2 Oxygen (O) atoms
2. Products:
- [tex]\(CO_2\)[/tex]: 1 Carbon (C) atom, 2 Oxygen (O) atoms
- [tex]\(H_2O\)[/tex]: 2 Hydrogen (H) atoms, 1 Oxygen (O) atom
### Step 2: Balance the Carbon Atoms
There are 2 Carbon atoms on the reactant side (from [tex]\(C_2H_6\)[/tex]) and only 1 Carbon atom in each [tex]\(CO_2\)[/tex] molecule on the product side.
To balance the Carbons, we need 2 [tex]\(CO_2\)[/tex] molecules.
[tex]\[C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O\][/tex]
### Step 3: Balance the Hydrogen Atoms
There are 6 Hydrogen atoms on the reactant side (from [tex]\(C_2H_6\)[/tex]) and 2 Hydrogen atoms in each [tex]\(H_2O\)[/tex] molecule on the product side.
To balance the Hydrogens, we need 3 [tex]\(H_2O\)[/tex] molecules.
[tex]\[C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O\][/tex]
### Step 4: Balance the Oxygen Atoms
Now we check the number of Oxygen atoms:
- On the product side, we have [tex]\(2 \times 2 = 4\)[/tex] Oxygen atoms from 2 [tex]\(CO_2\)[/tex] molecules, and [tex]\(3 \times 1 = 3\)[/tex] Oxygen atoms from 3 [tex]\(H_2O\)[/tex] molecules, giving a total of [tex]\(4 + 3 = 7\)[/tex] Oxygen atoms.
- On the reactant side, Oxygen comes from [tex]\(O_2\)[/tex] molecules, each providing 2 Oxygen atoms.
To balance this, we need [tex]\(7 / 2 = 3.5\)[/tex] [tex]\(O_2\)[/tex] molecules, which may not appear intuitive since half molecules are not usually used in balanced equations. We will double all the coefficients to avoid fractional molecules:
[tex]\[2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\][/tex]
### Final Step: Verify the Balance
1. Reactants:
- [tex]\(2C_2H_6\)[/tex]: [tex]\(2 \times 2 = 4\)[/tex] Carbon atoms, [tex]\(2 \times 6 = 12\)[/tex] Hydrogen atoms
- [tex]\(7O_2\)[/tex]: [tex]\(7 \times 2 = 14\)[/tex] Oxygen atoms
2. Products:
- [tex]\(4CO_2\)[/tex]: [tex]\(4 \times 1 = 4\)[/tex] Carbon atoms, [tex]\(4 \times 2 = 8\)[/tex] Oxygen atoms
- [tex]\(6H_2O\)[/tex]: [tex]\(6 \times 2 = 12\)[/tex] Hydrogen atoms, [tex]\(6 \times 1 = 6\)[/tex] Oxygen atoms
- Total Oxygen atoms: [tex]\(8 + 6 = 14\)[/tex]
Both sides have 4 Carbon atoms, 12 Hydrogen atoms, and 14 Oxygen atoms, verifying that the equation is balanced:
[tex]\[2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\][/tex]
### Step 1: Count the Atoms
1. Reactants:
- [tex]\(C_2H_6\)[/tex]: 2 Carbon (C) atoms, 6 Hydrogen (H) atoms
- [tex]\(O_2\)[/tex]: 2 Oxygen (O) atoms
2. Products:
- [tex]\(CO_2\)[/tex]: 1 Carbon (C) atom, 2 Oxygen (O) atoms
- [tex]\(H_2O\)[/tex]: 2 Hydrogen (H) atoms, 1 Oxygen (O) atom
### Step 2: Balance the Carbon Atoms
There are 2 Carbon atoms on the reactant side (from [tex]\(C_2H_6\)[/tex]) and only 1 Carbon atom in each [tex]\(CO_2\)[/tex] molecule on the product side.
To balance the Carbons, we need 2 [tex]\(CO_2\)[/tex] molecules.
[tex]\[C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O\][/tex]
### Step 3: Balance the Hydrogen Atoms
There are 6 Hydrogen atoms on the reactant side (from [tex]\(C_2H_6\)[/tex]) and 2 Hydrogen atoms in each [tex]\(H_2O\)[/tex] molecule on the product side.
To balance the Hydrogens, we need 3 [tex]\(H_2O\)[/tex] molecules.
[tex]\[C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O\][/tex]
### Step 4: Balance the Oxygen Atoms
Now we check the number of Oxygen atoms:
- On the product side, we have [tex]\(2 \times 2 = 4\)[/tex] Oxygen atoms from 2 [tex]\(CO_2\)[/tex] molecules, and [tex]\(3 \times 1 = 3\)[/tex] Oxygen atoms from 3 [tex]\(H_2O\)[/tex] molecules, giving a total of [tex]\(4 + 3 = 7\)[/tex] Oxygen atoms.
- On the reactant side, Oxygen comes from [tex]\(O_2\)[/tex] molecules, each providing 2 Oxygen atoms.
To balance this, we need [tex]\(7 / 2 = 3.5\)[/tex] [tex]\(O_2\)[/tex] molecules, which may not appear intuitive since half molecules are not usually used in balanced equations. We will double all the coefficients to avoid fractional molecules:
[tex]\[2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\][/tex]
### Final Step: Verify the Balance
1. Reactants:
- [tex]\(2C_2H_6\)[/tex]: [tex]\(2 \times 2 = 4\)[/tex] Carbon atoms, [tex]\(2 \times 6 = 12\)[/tex] Hydrogen atoms
- [tex]\(7O_2\)[/tex]: [tex]\(7 \times 2 = 14\)[/tex] Oxygen atoms
2. Products:
- [tex]\(4CO_2\)[/tex]: [tex]\(4 \times 1 = 4\)[/tex] Carbon atoms, [tex]\(4 \times 2 = 8\)[/tex] Oxygen atoms
- [tex]\(6H_2O\)[/tex]: [tex]\(6 \times 2 = 12\)[/tex] Hydrogen atoms, [tex]\(6 \times 1 = 6\)[/tex] Oxygen atoms
- Total Oxygen atoms: [tex]\(8 + 6 = 14\)[/tex]
Both sides have 4 Carbon atoms, 12 Hydrogen atoms, and 14 Oxygen atoms, verifying that the equation is balanced:
[tex]\[2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\][/tex]
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