Question

Enter your answers in scientific notation.

Carbon dioxide is removed from the atmosphere of space capsules by reaction with a solid metal hydroxide. The products are water and the metal carbonate.

(a) Calculate the mass of CO₂ that can be removed by reaction with 5.26 kg of lithium hydroxide.

_____g CO₂

(b) How many grams of CO₂ can be removed by 0.11 kg of each of the following: lithium hydroxide, magnesium hydroxide, and aluminum hydroxide?

By lithium hydroxide:

____g CO₂

By magnesium hydroxide:

_____g CO₂

By aluminum hydroxide:

_____g CO₂

Answer 1

The metal hydroxide is LiOH

The reaction is

2LiOH + CO₂ ---> Li₂CO₃ + H2O

a) as per given equation 2moles of LiOH (2 X23.95 g) can remove 1 mole of CO2 (44grams)

The mass of LiOH = 5.26Kg = 5260grams

So moles of LiOH = 5260 / 23.95 = 219.62 moles

It will remove moles of CO2 = 0.5 X 219.62 = 109.81 moles

Mass of CO₂ = Moles X molecular weight = 109.81 X 44 = 4831.64 grams of CO2

b)

(i)The reaction stoichiometry will be the same

2moles of LiOH (2 X23.95 g) can remove 1 mole of CO2 (44grams)

The mass of LiOH = 0.11Kg = 110 grams

So moles of LiOH = 110 / 23.95 = 4.59 moles

It will remove moles of CO2 = 0.5 X 4.59 = 2.295 moles

Mass of CO2 removed = Moles X molecular weight = 2.295 x 44 = 100.98 grams

(ii) The reaction of Mg(OH)2 with CO2 will be

Mg(OH)₂+ CO₂ ---> MgCO₃ + H2O

One mole of Mg(OH)2 will react with one mole of CO2

The mass of Mg(OH)2 = 0.11Kg = 110 grams

So moles of LiOH = 110 / 58.32 = 1.89moles

It will remove moles of CO2 = 1.89

Mass of CO2 that can be removed = 1.89 X 44 grams = 83.16 grams

(iii) The reaction of Al(OH)3 with CO2 will be

2Al(OH)₃+ 3CO₂ ---> Al₂(CO₃)₃ + 3H₂O

two moles of Al(OH)₃ will react with three moles of CO2

The mass of Al(OH)₃ = 0.11Kg = 110 grams

So moles of Al(OH)₃ = 110 / 78 = 1.41moles

It will remove moles of CO2 = 1.5 X 1.41 = 2.115 moles

Mass of CO2 that can be removed = 2.115 X 44 grams = 93.06 grams