# Calculate how many kg of lithium hydoxide must be stored on board for each day that the five-member crew is in flight? Help Im so confused???????????? Five member crew usually consumes 4 kg of oxygen everyday.  Cellular respiration formula: c6h12o6+6o2----> 6co2+6h2o second equation: co2+2LiOH---> Li2Co3+H2o

Once the oxygen is used by the crew `CO_2` is formed in the flight. `CO_2` is hazardous for health.So we need to remove that `CO_2` . So what is done here is forming Lithium carbonate using `CO_2` . It will remove `CO_2` from the air in flight and make a healthy environment.

`C_6H_12O_6+6O_2rarr 6CO_2+6H_2O`

Molar mass of Oxygen = 16g/mol

Amount of `O_2` consumed per day = 4kg

Amount of` O_2` moles consumed per day = `4000/32 ` = 125mol

Mole ratio

`O_2:CO_2 = 1:1`

Therefore;

Amount of `CO_2` formed per day = 125mol

`CO_2+2LiOH rarr Li_2CO_3+H_2O`

Mole ratio

`CO_2:LiOH = 1:2`

Therefore;

Amount of `LiOH` required per day = 125*2 = 250mol

Molar mass of `LiOH` = 24g/mol

mass of `LiOH` required `= 24*250 = 6000g = 6kg`

So for the process we need 6kg of LiOH.

Assumptions

• All reactions are complete reactions
• Only these two reactions are takes place in the system between the reactants

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