What mass of water is needed to react with 32.9 g lithium nitride

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.