The balanced chemical equation for the combustion of methane is: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) Which of the following statements concerning this chemical equation is/are correct? 1. One gram of methane gas reacts with two grams of dioxygen gas, producing one gram of carbon dioxide gas and two grams of gaseous water. 2. One mole of methane gas reacts with two moles of dioxygen gas, producing one mole of carbon dioxide gas and two moles of gaseous water. 3. One molecule of methane gas reacts with two molecules of dioxygen gas, producing one molecule of carbon dioxide gas and two molecules of gaseous water.

Option 2. One mole of methane gas reacts with two moles of dioxygen gas, producing one mole of carbon dioxide gas and two moles of gaseous water.

Explanation:

The clue is in the stoichiometry of the reaction:

1 CH₄(g) + 2 O₂(g) → 1 CO₂(g) + 2 H₂O(g)

In any chemistry reaction, the stoichiometric coefficients are moles; moles of reactants to produce moles of products

h let me think

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 :