Answer:
We can use stoichiometry to determine the amount of water produced when 29.3 g of LiOH reacts with CO2.
First, we need to convert 29.3 g of LiOH to moles:
moles of LiOH = mass/molar mass = 29.3 g / (6.941 g/mol + 15.999 g/mol + 1.008 g/mol) = 0.5 mol
From the balanced chemical equation, we see that 2 moles of LiOH react with 1 mole of CO2 to produce 1 mole of H2O. Therefore, we can say:
0.5 mol LiOH × (1 mol H2O / 2 mol LiOH) = 0.25 mol H2O
Now, we can use the molar mass of water to convert moles to grams:
mass of H2O = moles × molar mass = 0.25 mol × 18.015 g/mol = 4.504 g
Finally, we can use the density of water to convert grams to milliliters:
volume of H2O = mass / density = 4.504 g / 0.997 g/mL = 4.52 mL
Therefore, approximately 4.52 mL of water could form from 29.3 g of LiOH.