Determine the mass of lithium hydroxide produced when 0.38 grams of Li3 N reacts with water according to the following unbalanced chemical equation: ___Li3 N (s) + ___H2 O (l) → ___NH3 + ___LiOH

First, we need to balance the chemical equation. The balanced equation is:

Li3N (s) + 3H2O (l) → NH3 (g) + 3LiOH (s)

From the balanced equation, we can see that one mole of Li3N reacts with three moles of water to produce one mole of NH3 and three moles of LiOH.

Next, we need to find the molar mass of Li3N and LiOH.

The molar mass of Li3N is approximately 34.83 g/mol (6.94 g/mol for Li * 3 + 14.01 g/mol for N).

The molar mass of LiOH is approximately 23.95 g/mol (6.94 g/mol for Li + 15.999 g/mol for O + 1.008 g/mol for H).

Now, we can find the number of moles of Li3N in 0.38 grams by using the formula:

Number of moles = mass / molar mass

Number of moles of Li3N = 0.38 g / 34.83 g/mol = 0.0109 mol

Since the balanced equation shows that one mole of Li3N produces three moles of LiOH, we can find the number of moles of LiOH produced:

Number of moles of LiOH = 0.0109 mol * 3 = 0.0327 mol

Finally, we can find the mass of LiOH produced by multiplying the number of moles by the molar mass:

Mass of LiOH = number of moles * molar mass

Mass of LiOH = 0.0327 mol * 23.95 g/mol = 0.783 g

So, the mass of lithium hydroxide produced when 0.38 grams of Li3N reacts with water is approximately 0.783 grams.