The given chemical equation is:
HI + LiOH -> LiI + H2O
According to the equation, 1 mole of HI reacts with 1 mole of LiOH to produce 1 mole of LiI and 1 mole of water (H2O).
Given that 3 moles of water (H2O) are produced, we can say that 3 moles of LiI will also be produced. This is because the stoichiometry of the reaction tells us that 1 mole of LiI is produced for every 1 mole of water (H2O) produced.
Now, we need to calculate the mass of 3 moles of LiI.
The molar mass of LiI can be calculated as follows:
LiI = Li + I
LiI = 6.941 + 126.90
LiI = 133.84 g/mol
Therefore, the mass of 3 moles of LiI is:
Mass = 3 moles x 133.84 g/mol
Mass = 401.52 g
Hence, 401.52 grams of LiI will be produced when 3 moles of water (H2O) are produced.