Methane CH4 gas and oxygen O2 gas react to form carbon dioxide CO2 gas and water H2O vapor. Suppose you have 11.0 mol of CH4 and 7.0 mol of O2 in a reactor. Suppose as much as possible of the CH4 reacts. How much will be left? Round your answer to the nearest 0.1 mol

The number of moles of CH₄ that will remain is 7.5 moles

Explanation:

Complete combustion reaction of oxygen and methane is given as;

CH₄ + 2O₂ → CO₂ + 2H₂O

comparing the number of moles of methane and oxygen in the reaction above, the amount of methane used in the reaction is calculated as follows;

2 moles of O₂   > 1 mole of CH₄  

7 moles of O₂ > x moles of CH₄

x = 7 / 2

x = 3.5 moles of CH₄

Total moles of CH₄ in the reactor = 11.0 moles

The number of moles that will remain = 11.0 moles - 3.5 moles

The number of moles that will remain = 7.5 moles

Therefore, the number of moles of CH₄ that will remain is 7.5 moles

fe 55.85 g/mol and h2o 18.02 g/mol.

the maximum mass of nh3 is nh3 in grams