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Chapter 8: Problem 83

Air contains \(20 \% \mathrm{O}_{2}\), by volume. What volume of air is needed at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) for complete combustion of \(80 \mathrm{~g}\) methane? (a) 101 (b) 501 (c) 2241 (d) 11201

Short Answer

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Based on the step by step solution provided above, create a short answer: The volume of air required for the complete combustion of 80g of methane at 0掳C and 1 atm is 1120.5 L.

Step by step solution

01

Write the balanced chemical equation for the combustion of methane.

The balanced chemical equation for the combustion of methane is: CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (g)
02

Find the number of moles of methane.

We are given 80g of methane and the molar mass of methane (CH4) is 16g/mol. To find the number of moles (n) of methane, we will use the formula: n = mass of substance / molar mass of substance n (methane) = 80g / 16g/mol = 5 mol of CH4
03

Calculate the volume of oxygen required for combustion.

From the balanced chemical equation, 1 mol of CH4 requires 2 mol of O2 for complete combustion. So, for 5 mol of CH4, we need: 5 mol CH4 * (2 mol O2 / 1 mol CH4) = 10 mol O2 At 0掳C and 1 atm, we can use the ideal gas law to calculate the volume of oxygen required. PV = nRT where P = pressure (1 atm), n = moles of O2 (10 mol), R = gas constant (0.0821 L atm/mol K), and T = temperature (273K), V = volume of O2 V_O2 = nRT / P V_O2 = (10 mol * 0.0821 L atm/mol K * 273 K) / 1 atm = 224.1 L
04

Calculate the volume of air required for complete combustion.

Air contains 20% O2 by volume, so to obtain 224.1 L of O2, we need to find out the volume of air required. Let V_air be the volume of air required: (V_O2 / V_air) = 20% V_O2 = 0.20 * V_air V_air = V_O2 / 0.20 V_air = 224.1 L / 0.20 = 1120.5 L The volume of air required for complete combustion of 80g of methane is 1120.5 L. The answer is close to option (d) 11201, but there might be a typo in the given options. Therefore, the most accurate answer is 1120.5 L.

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