On the space shuttle, lithium hydroxide is produced from lithium oxide and water. the lithium hydroxide then reacts with the exhaled carbon dioxide to remove it from the air supply. the process occurs based on the following reactions: li2o (s) + h2o (g) ĺ2lioh (s) lioh (s) + co2 (g) ĺlihco3 (s) if you have 2.6 g of li2o and 1.3 g of h2o, how many grams of co2 can you remove from the air? report your answer to two significant figures. do not include units in your answer. __1__ g co2

6,2g of CO₂

Explanation:

Based on the reactions:

Li₂O(s) + H₂O(g) → 2LiOH(s)

LiOH(s) + CO₂(g) → LiHCO₃(s)

2,6g of Li₂O and 1,3g of H₂O are:

2,6g × ( 1mol / 29,88g) = 0,087 moles

1.3g × ( 1mol / 18,01g) = 0,072 moles

That means limiting reactant is H₂O. The moles produced of LiOH are:

0,072 moles of H₂O × ( 2mol LiOH / 1mol H₂O) = 0,14 moles of LiOH

Thus, the maximum CO₂ that can react are 0,14 moles of CO₂, in grams

0,14 moles CO₂ × (44,01g / 1mol) = 6,2g of CO₂

I hope it helps!

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

the answer is 0.42 is that is an option. this question provides no info