The combustion of propane (c3h8) produces co2 and h2o: c3h8 (g) + 5o2 (g) → 3co2 (g) + 4h2o (g) the reaction of 2.5 mol of o2 with 4.6 mol of c3h8 will produce mol of h2o.

2mol of with 4.6 mol  will produce 2mol .

Explanation:

The given information is:

2mol of with 4.6 mol of   will produce.....mol of

As we can see hear, 5mol of produce 4 mols of

Hence, According to the given equation above,

Mols of produced will be,

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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 :