When the following equation is balanced, the coefficients are:

[tex]\[ C_6H_{14}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g) \][/tex]

A. [tex]\(1, 9, 6, 7\)[/tex]
B. [tex]\(1, 1, 2, 10, 12\)[/tex]
C. [tex]\(2, 19, 12, 14\)[/tex]
D. [tex]\(2, 7, 6, 8\)[/tex]
E. [tex]\(2, 9, 4, 8\)[/tex]



Answer :

Let's balance the chemical equation step-by-step to determine the correct coefficients.

We start with the unbalanced equation:

[tex]\[ C_6H_{14}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g) \][/tex]

First, we need to balance the carbon (C) atoms. In [tex]\( C_6H_{14} \)[/tex], there are 6 carbon atoms. Therefore, we need 6 molecules of [tex]\( CO_2 \)[/tex]:

[tex]\[ C_6H_{14}(g) + O_2(g) \rightarrow 6 CO_2(g) + H_2O(g) \][/tex]

Next, let's balance the hydrogen (H) atoms. In [tex]\( C_6H_{14} \)[/tex], there are 14 hydrogen atoms. Therefore, we need 7 molecules of [tex]\( H_2O \)[/tex] (because each water molecule has 2 hydrogen atoms):

[tex]\[ C_6H_{14}(g) + O_2(g) \rightarrow 6 CO_2(g) + 7 H_2O(g) \][/tex]

Now, we'll balance the oxygen (O) atoms. On the right side, we have:

- [tex]\( 6 \times 2 = 12 \)[/tex] oxygen atoms from [tex]\( 6 CO_2 \)[/tex]
- [tex]\( 7 \times 1 = 7 \)[/tex] oxygen atoms from [tex]\( 7 H_2O \)[/tex]

That gives us a total of [tex]\( 12 + 7 = 19 \)[/tex] oxygen atoms on the right side. We need to balance these with the oxygen atoms on the left side. Each [tex]\( O_2 \)[/tex] molecule has 2 oxygen atoms, so we need [tex]\( \frac{19}{2} \)[/tex] or 9.5 [tex]\( O_2 \)[/tex] molecules:

[tex]\[ C_6H_{14}(g) + 9.5 O_2(g) \rightarrow 6 CO_2(g) + 7 H_2O(g) \][/tex]

Since we generally avoid fractional coefficients in balanced chemical equations, we multiply every coefficient by 2 to obtain whole numbers:

[tex]\[ 2 C_6H_{14}(g) + 19 O_2(g) \rightarrow 12 CO_2(g) + 14 H_2O(g) \][/tex]

Thus, the coefficients are:
[tex]\[ 2, 19, 12, 14 \][/tex]

Given these balanced coefficients, the correct answer among the provided options is:
[tex]\[ 2, 19, 12, 14 \][/tex]

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