29. Consider the following chemical equations and their enthalpy changes.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = −8.0 × 102 kJ
CaO(s) + H2O(_) → Ca(OH)2(aq) ΔH = −65 kJ
What volume of methane, at 20°C and 100 kPa, would have to be combusted in order to release the same amount of energy as the reaction of 1.0 × 102 g of CaO with sufficient water? (The volume of 1.00 mol of any gas at 20°C and 100 kPa is 24 L.)
i tried solving this question, i found the amount of energy it takes for the CaO and found the grams needed for methane to produce such energy, i get 2.3g but than how do i convert it to litres is confusing.... please help.
Can't you use PV = nRT?
To solve this problem, you need to determine the number of moles of methane required to release the same amount of energy as the reaction of 1.0 x 10^2 g of CaO. Then you can convert the moles of methane gas to liters.
Let's break down the steps:
Step 1: Calculate the moles of CaO
Use the molar mass of CaO to calculate the number of moles. The molar mass of CaO is the sum of the atomic masses of calcium (Ca) and oxygen (O):
Ca: 40.08 g/mol
O: 16.00 g/mol
Molar mass of CaO = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol
Now we can calculate the number of moles of CaO:
Moles of CaO = Mass of CaO / Molar mass of CaO
Moles of CaO = 1.0 x 10^2 g / 56.08 g/mol
Moles of CaO ≈ 1.78 mol
Step 2: Calculate the moles of methane
By comparing the enthalpy changes in the given chemical equations, we see that the enthalpy change for the combustion of 1 mole of methane is -8.0 x 10^2 kJ.
So the moles of methane required to release the same amount of energy as the reaction of 1.78 mol of CaO can be calculated using the stoichiometry of the equation:
1 mole of CH4(g) + 2 moles of O2(g) → 1 mole of CO2(g) + 2 moles of H2O(g)
Since the coefficient of CH4 is 1, the number of moles of CH4 is equal to the number of moles of CaO:
Moles of CH4 = Moles of CaO ≈ 1.78 mol
Step 3: Convert moles of methane to liters
Given that the molar volume of any gas at 20°C and 100 kPa is 24 L/mol, we can convert the moles of CH4 to liters:
Volume of CH4 = Moles of CH4 x Molar volume
Volume of CH4 = 1.78 mol x 24 L/mol
Volume of CH4 ≈ 42.7 L
Therefore, the volume of methane required to release the same amount of energy as the reaction of 1.0 x 10^2 g of CaO is approximately 42.7 liters.