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What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, [tex]C_2H_6[/tex]?

In this reaction, ethane is burned in the presence of oxygen [tex](O_2)[/tex] to form carbon dioxide [tex](CO_2)[/tex] and water [tex](H_2O)[/tex].

[tex]\[ C_2H_6(g) + \ldots O_2(g) \rightarrow \ldots CO_2(g) + \ldots H_2O(g) \][/tex]

Recall that the coefficients of the final balanced equation should be whole numbers. Thus, you might need to multiply through the equation by a factor of two to obtain whole numbers in your last step.

Express the coefficients as integers separated by commas.

Sagot :

GinnyAnswer
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
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