2. How many grams of methane must be burned to produce 2.05 grams of water in the complete combustion reaction? The unbalanced equation for the reaction is: _CH418) + _ 0218) ►_CO2(8) +_ H2O(8)

0.88 g

Explanation:

The balanced reaction equation is;

CH4(g) + 2O2(g) > CO2(g) + 2H2O(l)

Number of moles in 2.05g of water = 2.05g/18g/mol = 0.11 moles

If 1 mole of CH4 yields 2 moles of water

x moles of CH4 yields 0.11 moles of water

2x = 0.11

x = 0.11/2

x= 0.055 moles

Molar mass of methane = 16 g/mol

Mass of methane = 16 g/mol * 0.055 moles

Mass of methane =0.88 g

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 :

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