During combustion, methine yeilds carbon dioxide and water. the unbalanced equation for this reaction is : CH4(g)+O2(g)>CO2(g)+H2O(i)

what will the mole ratios for the balanced equation be?(what coefficients are needed in order to balanced the equation?)

CH4 + 2O2 --> CO2 + 2H2O

To balance the equation, we need to determine the coefficients that will ensure that the same number of atoms of each element are present on both sides of the equation.

Let's balance the equation step by step:

1. Start by balancing the carbon atoms. There is one carbon atom in methane (CH4) and one carbon atom in carbon dioxide (CO2). Therefore, we need to place a coefficient of 1 in front of CO2.

CH4(g) + O2(g) > 1CO2(g) + H2O(i)

2. Next, balance the hydrogen atoms. There are four hydrogen atoms in methane and two in water. To balance the hydrogen, we need a coefficient of 2 in front of H2O.

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

3. Balancing the oxygen atoms is the last step. On the left side, we have two oxygen atoms from O2, and on the right side, we have two from CO2 and four from H2O. This gives us a total of six oxygen atoms on the right side. To balance the oxygen, we need a coefficient of 3/2 in front of O2, which will give us three oxygen atoms.

CH4(g) + (3/2)O2(g) > 1CO2(g) + 2H2O(i)

To avoid fractional coefficients, we can multiply all coefficients by 2 to obtain the final balanced equation:

2CH4(g) + 3O2(g) > 2CO2(g) + 4H2O(i)

So, the mole ratios for the balanced equation are:

2 moles of methane reacts with 3 moles of oxygen gas to produce 2 moles of carbon dioxide and 4 moles of water.