Enter your answer in the provided box. One of the air-purification methods for e
ID: 1027574 • Letter: E
Question
Enter your answer in the provided box. One of the air-purification methods for enclosed spaces involves the use of "scrubbers" containing aqueous lithium hydroxide, which reacts with carbon dioxide to produce lithium carbonate and water: 2 LiOH(aq) + 2 CO2(g) Li2CO3(s) + H20() Consider the air supply in a submarine with a total volume of 1.85 x 10 L. The pressure is 0.9980 atm, and the temperature is 25°C. If the pressure in the submarine drops by 0.0100 atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many CO2 are consumed? moles of molExplanation / Answer
First calculate mole of CO2 at 0.9980 atm pressure by using ideal gas equation
Ideal gas equation
PV = nRT where, P = atm pressure = 0.9980 atm,
V = volume in Liter = 1.85 X 105 L
n = number of mole = ?
R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = 250C = 273.15+ 25 = 298.15 K
We can write ideal gas equation
n = PV/RT
Substitute the value
n = (0.9980 X 1.85 X 105)/(0.08205 X 298.15) = 7547.25 mole
initialy present 7547.25 mole of CO2
now calculate mole of CO2 at 0.0100 atm pressure by using ideal gas equation
Ideal gas equation
PV = nRT where, P = atm pressure = 0.0100 atm,
V = volume in Liter = 1.85 X 105 L
n = number of mole = ?
R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,
T = Temperature in K = 250C = 273.15+ 25 = 298.15 K
We can write ideal gas equation
n = PV/RT
Substitute the value
n = (0.0.0100 X 1.85 X 105)/(0.08205 X 298.15) = 75.62 mole
finaly present 75.62 mole of CO2
no. of mole consumed = no. of mole initialy present - no. of mole remain after reaction
no. of mole of CO2 consumed = 7547.25 - 75.62 = 7471.63 mole
7471.63 mole of CO2 consumed.
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