   # The space shuttle environmental control system handles excess CO 2 (which the astronauts breathe out; it is 4.0% by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, Li 2 CO 3 , and water. If there are seven astronauts on board the shuttle, and each exhales 20. L of air pee minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 5, Problem 114E
Textbook Problem
510 views

## The space shuttle environmental control system handles excess CO2 (which the astronauts breathe out; it is 4.0% by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, Li2CO3, and water. If there are seven astronauts on board the shuttle, and each exhales 20. L of air pee minute, how long could clean air be generated if there were 25,000 g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 g/mL.

Interpretation Introduction

Interpretation: The time required to clean the air if there were 25000 g of LiOH pellets available for each shuttle mission is to be calculated.

Concept introduction: The number of moles is defined as the ratio of mass with the molecular mass of an element. The mass of an element is the amount of the substance present in an element. Density is defined as the ratio of mass per unit volume.

To determine: The time required to clean the air if there were 25000 g of LiOH pellets available for each shuttle mission.

### Explanation of Solution

Explanation

Given

The space shuttle environmental control system handles excess CO2 which the astronauts breathe out.

The excess CO2 is 4.0% by mass of exhaled air.

Density of air is 0.0010 g/mL .

Volume of exhaled CO2 is 20. L/min .

The conversion of L to mL is done as,

1 L=1000 mL

Therefore, the 20L to mL is done as,

20 L=20×1000 mL

The balanced chemical equation is,

CO2+2LiOHLi2CO3+H2O

The moles of LiOH is calculated by using the formula,

Numberofmoles =Mass Molarmass

The molar mass of LiOH is 23.9 g/mol .

Substitute the value of Mass and Molar mass of LiOH in the above expression.

Numberofmoles of LiOH=Mass of LiOHMolarmass of LiOH  =25000 g23.9 g/mol=1046.02 mol

Therefore, the moles of LiOH is 1046.02 mol .

The mass of CO2 is calculated by using the expression.

Mass=Density×Volume

Substitute the values of density and mass in the above expression to calculate the mass of CO2 .

Mass=Density×Volume=0.0010 g/mL×20×1000 mL/min=20 g/min

The moles of CO2 available (breathe out by astronaut) is calculated as by using the expression.

Moles of CO2=Mass of CO2Molar Mass of CO2

Substitute the values of mass and molar mass of CO2 to calculate the moles of CO2 available (breathe out by 7 astronauts) in the above expression

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