Lithium Nitride reacts with water to form ammonia and aqueous lithium hydroxide.

Li3 N(s) + 3H2 O(l) -----> NH3(g) + 3LIOH(aq)

a. What mass of water is needed to react with 32.9g Li3 N?
b. When the above reaction takes place, how many molecules of NH3 are pruduced?
c. Calculate the number of grams of Li3 N that must be added to an excess of water to produce 15.0L NH3 (at STP).

a. You have 32.9g of Li3N... WHat is this is moles?
Water reacts with Li3N on a 3:1 ratio. Therefore the value you got for part a, multiply by 3 to get the moles for water. Turn this value into grams.

... All this assuming you know the moles/molecular mass/weight in grams triangle.

b. I don't understand this question... the equation is already balanced is it not?

a) First off you have to figure out how many moles of Lithium Nitride is reacting. 32.9g/(34.8g/mole) This gives you .945 moles of Li3N. So you need 3 times as many moles of H2O to complete the reaction. . 945 X 3 = 2.84 moles H2O. Then you need to convert that to moles, 2.84 moles X 18.0 g/mole = 51.1 g H2O.

b) This is simply an Avagadro's number usage. There are 6.02 X 10^23 molecules in 1 mole of a substance. So you take the 1 mole X 6.02 X 10^23molecules/mole = 6.02 X 10^23 molecules NH3.

c) This question is a multiple application problem. First you need to convert the 15.0L NH3 to moles. This is done using the conversion factor of 22.4L per mole of gas @ STP. 15.0L/(22.4L/mole)= .670 moles NH3. Then, since the mole ratio of Li3N consumed to NH3 produced is 1:1, you need the same amount of Li3N, or .670 moles of Li3N. Then you just convert to grams. . 670 moles Li3N X 34.8g/mole=23.3 g Li3N.