kgr7lRosaag5aile
contestada

The carbon dioxide exhaled by astronauts can be removed from a spacecraft by reacting it with lithium hydroxide (LiOH). The reaction is as follows: CO2 (g) + 2LiOH (s) Li2CO3 (s) + H2O (l). An average person exhales about 20 moles of CO2 per day. How many moles of LiOH would be required to maintain two astronauts in a spacecraft for three days?

Respuesta :

meerkat18
Two astronauts would exhale about 40 moles of carbon dioxide daily.

Carbon dioxide reacts with lithium hydroxide in a 1 : 2 mole ratio. Set up a proportion:

 1 : 2 = 40 : x

Then, find x: 12=40x

 Cross multiply. x = 80 moles of LiOH per day for both astronauts
Eduard22sly

The number of mole of LiOH required to maintain two astronauts in the spacecraft for three days is 240 moles

How to determine the mole of LiOH required to react with 20 moles of CO₂ per day

CO₂ + 2LiOH —> Li₂CO₃ + H₂O

From the balanced equation above,

1 mole of CO₂  required 2 moles of LiOH.

Therefore,

20 moles of CO₂ will require = 20 × 2 = 40 moles of LiOH

How to determine the mole of LiOH required by two astronauts for 3 days

  • Mole of LiOH for 1 person per day = 40 moles
  • Mole of LiOH for 2 persons per day = 2 × 40 = 80 moles
  • Mole of LiOH for 2 persons for 3 days = 80 × 3 = 240 moles

Learn more about stoichiometry:

https://brainly.com/question/14735801

Otras preguntas