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To balance the chemical equation for the combustion of ethane ([tex]\(C_2H_6\)[/tex]), we need to ensure that there are the same number of each type of atom on both the reactant and product sides of the equation.
1. Write down the unbalanced equation:
[tex]\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \][/tex]
2. Balance carbon (C) atoms:
There are 2 carbon atoms in [tex]\(C_2H_6\)[/tex]. Therefore, we need 2 molecules of [tex]\(CO_2\)[/tex] to balance the carbon atoms:
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \][/tex]
3. Balance hydrogen (H) atoms:
There are 6 hydrogen atoms in [tex]\(C_2H_6\)[/tex]. Thus, we need 3 molecules of [tex]\(H_2O\)[/tex] to balance the hydrogen atoms:
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
4. Balance oxygen (O) atoms:
On the product side, we have a total of [tex]\(4\)[/tex] oxygen atoms from [tex]\(2CO_2\)[/tex] (since each [tex]\(CO_2\)[/tex] molecule has 2 oxygen atoms) and [tex]\(3\)[/tex] more oxygen atoms from [tex]\(3H_2O\)[/tex] (since each [tex]\(H_2O\)[/tex] molecule has 1 oxygen atom). That makes a total of [tex]\(4 + 3 = 7\)[/tex] oxygen atoms on the product side.
To balance the oxygen atoms on the reactant side, we need [tex]\(7/2\)[/tex] molecules of [tex]\(O_2\)[/tex] (since each molecule of [tex]\(O_2\)[/tex] contains 2 oxygen atoms):
[tex]\[ C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
5. Convert to whole numbers:
The coefficients must be whole numbers. To get rid of the fraction, multiply all coefficients by 2:
[tex]\[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \][/tex]
Now, the chemical equation is balanced, and the coefficients are:
[tex]\[ 2, 7, 4, 6 \][/tex]
These coefficients represent the stoichiometric amounts of each substance involved in the balanced combustion reaction of ethane.
So, the coefficients are as follows:
2, 7, 4, 6
1. Write down the unbalanced equation:
[tex]\[ C_2H_6 + O_2 \rightarrow CO_2 + H_2O \][/tex]
2. Balance carbon (C) atoms:
There are 2 carbon atoms in [tex]\(C_2H_6\)[/tex]. Therefore, we need 2 molecules of [tex]\(CO_2\)[/tex] to balance the carbon atoms:
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + H_2O \][/tex]
3. Balance hydrogen (H) atoms:
There are 6 hydrogen atoms in [tex]\(C_2H_6\)[/tex]. Thus, we need 3 molecules of [tex]\(H_2O\)[/tex] to balance the hydrogen atoms:
[tex]\[ C_2H_6 + O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
4. Balance oxygen (O) atoms:
On the product side, we have a total of [tex]\(4\)[/tex] oxygen atoms from [tex]\(2CO_2\)[/tex] (since each [tex]\(CO_2\)[/tex] molecule has 2 oxygen atoms) and [tex]\(3\)[/tex] more oxygen atoms from [tex]\(3H_2O\)[/tex] (since each [tex]\(H_2O\)[/tex] molecule has 1 oxygen atom). That makes a total of [tex]\(4 + 3 = 7\)[/tex] oxygen atoms on the product side.
To balance the oxygen atoms on the reactant side, we need [tex]\(7/2\)[/tex] molecules of [tex]\(O_2\)[/tex] (since each molecule of [tex]\(O_2\)[/tex] contains 2 oxygen atoms):
[tex]\[ C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \][/tex]
5. Convert to whole numbers:
The coefficients must be whole numbers. To get rid of the fraction, multiply all coefficients by 2:
[tex]\[ 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \][/tex]
Now, the chemical equation is balanced, and the coefficients are:
[tex]\[ 2, 7, 4, 6 \][/tex]
These coefficients represent the stoichiometric amounts of each substance involved in the balanced combustion reaction of ethane.
So, the coefficients are as follows:
2, 7, 4, 6
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