How many grams of li2o must be carried on board to remove 75.0kg of water?

To determine the amount of lithium oxide (Li2O) required to remove 75.0 kg of water, we first need to know the stoichiometry of the reaction involved in water removal. Typically, Li2O reacts with water to form lithium hydroxide (LiOH). The reaction is:

[ \text{Li}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{LiOH} ]

From the balanced equation, 1 mole of Li2O reacts with 1 mole of water. The molar mass of Li2O is approximately 29.88 g/mol. Thus, for 75.0 kg (or 75,000 g) of water, you need around 2,513.4 g of Li2O, calculated using the molar ratiOS and the respective molar masses.