To determine the molar ratio of ethane (C₂H₆) to water (H₂O) from the balanced chemical equation, let's carefully analyze the coefficients of each compound involved in the reaction.
The balanced chemical equation for the combustion of ethane is:
[tex]\[ 2 \, C_2H_6(g) + 7 \, O_2(g) \rightarrow 4 \, CO_2(g) + 6 \, H_2O(g) \][/tex]
From this equation, we can identify the coefficients of ethane (C₂H₆) and water (H₂O):
- The coefficient for ethane (C₂H₆) is 2.
- The coefficient for water (H₂O) is 6.
To find the molar ratio of ethane to water, we use these coefficients directly:
[tex]\[ \text{Molar ratio} = \frac{\text{Coefficient of ethane}}{\text{Coefficient of water}} = \frac{2}{6} \][/tex]
Simplifying this fraction, we get:
[tex]\[ \frac{2}{6} = \frac{1}{3} \][/tex]
Thus, the molar ratio of ethane to water is 1:3.
So, the correct answer is:
[tex]\[ 1:3 \][/tex]
This ratio corresponds to the third option in the provided choices:
[tex]\[ \boxed{1 : 3} \][/tex]
