How many moles of methane gas are required to produce 9 moles of water when it combusts in the presence of excess oxygen gas. The other product is carbon dioxide

The answer to your question is 4.5 moles of CH₄ are necessary to produce 9 moles of water.

Explanation:

Data

moles of methane = ?

moles of water = 9

moles of oxygen = excess

Balanced chemical reaction

               1 CH₄  +  2O₂  ⇒  CO₂  + 2H₂O

Process

1.- Use proportions to find the moles of Methane and consider the coefficients of the balanced chemical reaction.

              1 mol of CH₄ 2 moles of H₂O

                x                   9 moles of H₂O

                x = (9 x 1) / 2

                x = 9/2

                x = 4.5 moles of CH₄

the millers tale is the answer.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :