Answer to Question #173639 in General Chemistry for Ayesha

Question

Lithium hydroxide can be used to remove carbon dioxide exhaled by astronauts. The reaction  involves lithium hydroxide reacting with carbon dioxide to produce lithium hydrogen carbonate.  Each cartridge that is used has a lithium hydroxide mass of 3.00 x 104 g. If the process for  removing carbon dioxide from the air has a percentage yield of 83.0 %, what mass of carbon  dioxide is removed by each canister of lithium hydroxide.

Answer 1

2LiOH + CO₂ $\to$ Li₂CO₃ + H₂O

Moles of Lithium Hydroxide = 3×10⁴/23.95 = 12.52×10²

2 moles of LiOH react with 1 mole of CO₂

So ,

Moles of Carbon dioxide = 6.26×10²

Mass of carbon dioxide = 6.26×10² × 44.01 = 2.75×10⁴ g

Percentage yield = (Actual yield/theoretical yield)×100

83 =( Actual yield / 2.75×10⁴ g)×100

Actual yield =( 2.75×10⁴ ×83) /100

= 2.28×10⁴

Removed = theoretical yield - actual yield

= 2.75×10⁴ --2.28×10⁴

= 4.7×10³ g