Write a balanced equation for the combustion of gaseous methane (CH4), a majority component of natural gas, in which it combines with gaseous oxygen to form gaseous carbon dioxide and gaseous water.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

Explanation:

Step 1: Data given

gaseous methane = CH4 (g)

Combustion reaction is adding O2. The products will be carbondioxide (CO2) and water vapor (H2O)

Step 2: The unbalanced equation

CH4 (g) + O2 (g) → CO2 (g) + H2O (g)

Step 3: Balancing the equation

CH4 (g) + O2 (g) → CO2 (g) + H2O (g)

On the left side we have 4x H (in CH4), on the right side we have 2x H (in H2O). To balance the amount H on both sides, we have to multiply H2O by 2.

CH4 (g) + O2 (g) → CO2 (g) + 2H2O (g)

On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in 2H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side, by 2. Now the equation is balanced.

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

CH₄ (g) + 2O₂ (g) - --> 1CO₂ (g) + 2H₂O (g)

Explanation:

any combustion of a hydrocarbon equation is in form:

CₓHₐ (g) + BO₂ (g) - --> YCO₂ (g) + ZH₂O (g), where x, a, b, y, z are all whole number positive integers

there will be 1 CO₂ to 2 H₂O, since there is 1 C to 4 H in CH₄; it is not 1:4 since 2 H is needed in H₂O

CH₄ (g) + _O₂ (g) - --> 1CO₂ + 2H₂O

there is 4 total O on products side, which can make 2O₂

CH₄ (g) + 2O₂ (g) - --> 1CO₂ (g) + 2H₂O (g)