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Chapter 12: Problem 25

How many liters of \(\mathrm{CH}_{4}(\mathrm{g}),\) measured at \(23.4^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg},\) must be burned to provide the heat needed to vaporize 3.78 L of water at \(100^{\circ} \mathrm{C}\) ? \(\Delta \mathrm{H}_{\text {combustion }}=\) \(-8.90 \times 10^{2} \mathrm{kJmol}^{-1} \mathrm{CH}_{4} \quad\) For \(\quad \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad\) at \(\quad 100^{\circ} \mathrm{C}\) \(d=0.958 \mathrm{g} \mathrm{cm}^{-3},\) and \(\Delta H_{\mathrm{vap}}=40.7 \mathrm{kJmol}^{-1}\)

Short Answer

Expert verified
So, about 225 liters of \(\mathrm{CH}_{4}(\mathrm{g})\) are needed to vaporize 3.78 liters of water at \(100^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate the moles of water

We start by finding the number of moles of water that we have. Since \(1 \mathrm{cm}^{3}=1 \mathrm{mL}\), and \(1 \mathrm{L}=1000 \mathrm{mL}\), we can convert the volume of water in liters to volume in cubic centimeters. Hence, 3.78 L of water equals 3.78 × 1000 = 3780 cm3 (or mL). From the density value given, 0.958 g of water has a volume of 1 cm3, so 3780 cm3 of water has mass = 0.958 g/cm3 x 3780 cm3 = 3621.64 g. We then use the molar mass of water (18.02 g/mol) to find the number of moles of water, which equals 3621.64 g ÷ 18.02 g/mol = 200.97 moles of water.
02

Calculate the heat for vaporization

To calculate the required heat for vaporization, we use the formula q = nΔH where q represents the heat, n the number of moles, and ΔH the heat of vaporization. Plugging in the values, the heat required = 200.97 moles x 40.7 kJ/mol = 8179.5 kJ.
03

Calculate the moles of methane

We want to find how much CH4 should be burned to provide this heat. Since the enthalpy of combustion ΔH = -890 kJ/mol, this means that burning 1 mol of CH4 generates 890 kJ. Using the formula q = nΔH where q represents the heat and η the number of moles, the number of moles of CH4 needed equals heat required ÷ absolute value of enthalpy of combustion = 8179.5 kJ ÷ 890 kJ/mol = 9.19 moles of CH4.
04

Calculate the volume of methane

To convert the number of moles of methane to volume, we can use the ideal gas law, PV=nRT. Rearranging to solve for V gives V = nRT/P. Given that R = 0.08206 L.atm/(mol.K), the temperature T = 23.4 + 273.15 = 296.55 K (converting Celsius to Kelvin), and the pressure P = 768 mmHg = 1.01 atm (converting mmHg to atm), the volume required = 9.19 mol x 0.08206 L.atm/(mol.K) x 296.55 K ÷ 1.01 atm = 224.7 L.

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