Answer:
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)
Answer:
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)
