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consider the following reaction
li3n (s) + 3h2o (l) --> nh3 (g) + 3lioh (l)
if you need to make 120 g lioh, how many grams of li3n must you react with excess water

-700g
-117g
-350g
-58g

The balanced chemical reaction is:

Li3N (s) + 3H2o (l) --> NH3 (g) + 3LiOh (l)

We are given the amount of LiOH produced from the reaction. This will be the starting point of the calculations. We do as follows:

120 g LiOH ( 1 mol / 23.95 g ) ( 1 mol Li3N / 3 mol LiOH ) ( 34.83 g / 1 mol ) = 58.2 g

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