In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Step by step solution

01

Writing the balanced chemical equation

The chemical equation for this reaction is \[\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) + 2\mathrm{CO}_{2}(\mathrm{g}) \rightarrow 2\mathrm{Na}_{2} \mathrm{CO}_{3}\left(\mathrm{s}\right) + \mathrm{O}_{2}(\mathrm{g})\] Remember to balance the equation, each side of the equation must have the same number of atoms of each element.

02

Calculate the amount of CO2 exhaled per minute

The volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min}\) and \(3.8 \% \) of this volume is CO2. So, the volume of CO2 exhaled per minute can be calculated as \(4.0 \mathrm{L/min} \times 0.038 = 0.152 \mathrm{L/min} = 152 \mathrm{mL/min}\).

03

Calculate the volume of O2 produced per minute

From the balanced chemical equation, it can be seen that 1 mole of O2 is produced from 2 moles of CO2. Therefore, the volume of O2 produced per minute is half the volume of CO2 exhaled per minute, which is \(152 / 2 = 76 \mathrm{mL/min}\).

04

Calculate the rate of Na2O2 consumed

The balanced equation also shows that 1 mole of Na2O2 reacts with 2 moles of CO2. Therefore, the moles of Na2O2 consumed per minute is half the moles of CO2 exhaled per minute. Using the molar volume of a gas at the given conditions (approximately 24.5 L/mol) and the for simplicity envision that \(5 \times 10^{-4} \) moles of CO2 are exhaled every minute. According to the equation, 1 mole of Na2O2 reacts with 2 moles of CO2, so the consumption rate of Na2O2 is \(2.5 \times 10^{-4} \) moles per minute. Multiplying this by the molar mass of Na2O2 (\(77.9778 \,g/mol\)) gives the consumption rate in grams per minute. This should be multiplied by 60 to get the consumption rate in grams per hour.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!