. Please help me with a chemistry question????

. The space shuttle envrionmental control system handles carbon dioxdie (4% of exhaled air) by reacting with Lithium Hydroxide (LiOH) pellets to form lithium carbonate (Li2CO3) and water. if there are 7 astronauts on board the shuttle, and each exhale 20L of air per minute, how long could clean air be generated if there were 25,000g of lithium hydroxide pellets available for each shuttle mission (round to the nearest 10 minutes)? Assume the density of air is 0.0010g/ml

1 Answer
Jun 17, 2018

Answer:

The #"LiOH"# could regenerate the air for 70 h.

Explanation:

Step 1. Calculate the mass of #"CO"_2# treated

#M_text(r): color(white)(mmm)23.95color(white)(mll)44.01#
#color(white)(mmmmll)"2LiOH" + "CO"_2 → "Li"_2"CO"_3 + "H"_2"O"#
#"Mass/g:"color(white)(m)"25 000"#

#m_text(CO₂) = "25 000" color(red)(cancel(color(black)("g LiOH"))) × (1 color(red)(cancel(color(black)("mol LiOH"))))/(23.95 color(red)(cancel(color(black)("g LiOH")))) × (1 color(red)(cancel(color(black)("mol CO"_2))))/(2 color(red)(cancel(color(black)("mol LiOH")))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "23 000 g CO"_2#

Step 2. Calculate the mass of air treated

#m_text(air) = "23 000"color(red)(cancel(color(black)( "g CO"_2))) × "100 g air"/(4 color(red)(cancel(color(black)("g CO"_2)))) = "570 000 g air"#

Step 3. Calculate the volume of air treated

#V_text(air) = "570 000" color(red)(cancel(color(black)("g air"))) × (1 color(red)(cancel(color(black)("mL air"))))/(0.0010 color(red)(cancel(color(black)("g air")))) × "1 L air"/(1000 color(red)(cancel(color(black)("mL air")))) = "570 000 L air"#

Step 4. Calculate the time required

#"Rale of air usage" = 7 color(red)(cancel(color(black)("astronauts"))) × "20 L air"/(1 color(red)(cancel(color(black)("atronaut")))·"min") = "140 L air/min"#

#t = "570 000" color(red)(cancel(color(black)("L"))) × "1 min"/(140 color(red)(cancel(color(black)("L air")))) = "4100 min = 70 h"#

Note: The answer can have only one significant figure because that is all you gave for the percentage of carbon dioxide.

The best we can do is estimate the time to the nearest 10 h.