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Chapter 8: Problem 84

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

Expert verified
(Assume 20% oxygen by volume in the air.) a) 240 L b) 2000 L c) 2241 L d) 2500 L Answer: c) 2241 L

Step by step solution

01

Write down the balanced combustion reaction for methane.

The combustion reaction for methane is: CH4 + 2O2 -> CO2 + 2H2O
02

Calculate the moles of methane (CH4) involved in the reaction.

To determine the number of moles of methane, use the formula: Number of moles = Mass / Molar mass Methane (CH4) has a molar mass of 12 (carbon) + 4(1) (hydrogen) = 16 g/mol. Number of moles of CH4 = 80 g / 16 g/mol = 5 moles
03

Calculate the moles of oxygen (O2) required for the complete combustion.

Based on the stoichiometry of the reaction, 2 moles of O2 are required for 1 mole of CH4. Therefore, for 5 moles of CH4, we need: 5 moles CH4 * (2 moles O2 / 1 mole CH4) = 10 moles O2
04

Calculate the total moles of air needed for complete combustion.

Since air contains only 20% O2 by volume, we need enough air that contains 10 moles of O2. Let x be the total moles of air: 0.20x = 10 moles O2 x = 50 moles (total moles of air)
05

Use the ideal gas law to calculate the volume of air needed.

Now we can use the ideal gas law, PV=nRT, to calculate the volume of air needed. We are given the pressure (P = 1 atm) and temperature (T = 0°C = 273.15 K). The ideal gas constant (R) is 0.0821 L atm / K mol. 1 atm * V = (50 moles) * (0.0821 L atm / K mol) * (273.15 K) V = 2241 L From the choices provided, the correct answer is (c) 2241.

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Most popular questions from this chapter

An organic compound contains \(40 \%\) carbon and \(6.67 \%\) hydrogen by mass. Which of the following represents the empirical formula of the compound? (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{2} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{O}\)

A compound having the empirical formula, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\), has a molecular weight of \(170 \pm 5 .\) The molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{3}\)

For \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\), which is the correct mole relationship? (a) \(9 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}\) (b) \(5 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}\) (c) \(9 \times\) mole of \(\mathrm{Cu}=\) mole of \(\mathrm{O}_{2}\) (d) mole of \(\mathrm{Cu}=5 \times\) mole of \(\mathrm{O}\)

A gaseous oxide contains \(30.4 \%\) of nitrogen, one molecule of which contains one nitrogen atom. The density of the oxide relative to oxygen, under identical conditions, is about (a) \(0.69\) (b) \(1.44\) (c) \(0.35\) (d) \(2.88\)

A quantity of \(0.2 \mathrm{~g}\) of an organic compound containing, \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\), on combustion yielded \(0.147 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.12 \mathrm{~g}\) water. The percentage of oxygen in it is (a) \(73.29 \%\) (b) \(78.45 \%\) (c) \(83.23 \%\) (d) \(89.50 \%\)
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