Respuesta :
Answer:
Combustion reaction
2C2H6+7O2 → 4O2+6H2O
Explanation:
In a combustion reaction with a hydrocarbon in the reactant side you will always have O2 as another reactant. As you will always have CO2 and H2O as the products.
Knowing that much you can set up your reaction equation..
C2H6+O2→ CO2+H2O
Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.
Start with the C atoms first and move to the H atoms next. It's easier to leave the O2 to the last, it has a way to alter the equation.
Initially, you would arrive at this, before the O2 has been balanced:
C2H6+O2→2CO2+3H2O
But, as you can see, you have an odd amount of O2 on the product side. In this case, you have to find the common factor of the amount of O on the product side and 2, Because of the O2 diatom. Therefore, 14 would be the lowest common factor of 2 and 7.
I hope it helped you!
Combustion reaction
2C2H6+7O2 → 4O2+6H2O
Explanation:
In a combustion reaction with a hydrocarbon in the reactant side you will always have O2 as another reactant. As you will always have CO2 and H2O as the products.
Knowing that much you can set up your reaction equation..
C2H6+O2→ CO2+H2O
Now the balancing can begin. Balancing hydrocarbon combustion reactions can be tricky, but if with practice they can be really fun and very rewarding.
Start with the C atoms first and move to the H atoms next. It's easier to leave the O2 to the last, it has a way to alter the equation.
Initially, you would arrive at this, before the O2 has been balanced:
C2H6+O2→2CO2+3H2O
But, as you can see, you have an odd amount of O2 on the product side. In this case, you have to find the common factor of the amount of O on the product side and 2, Because of the O2 diatom. Therefore, 14 would be the lowest common factor of 2 and 7.
I hope it helped you!
Answer: The chemical equation for the combustion of ethane is written below.
Explanation:
Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
[tex]\text{Hydrocarbon}+O_2\rightarrow CO_2+H_2O[/tex]
The chemical equation for the combustion of ethane follows:
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of ethane reacts with 7 moles of oxygen gas to produce 4 moles of carbon dioxide and 6 moles of water
Hence, the chemical equation for the combustion of ethane is written above.
