Consider the following reaction:
li3n (s) + 3h20 (1) ► nh3 (g) + 3lioh (1)
if you need to make 120 g lioh, how many grams of li3n must you react with
excess water?
700 g
350 g
o 117 g
58 g

Option D is correct =  58 g

Explanation:

Data Given:

mass of LiOH = 120 g

Mass of Li3N= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

                  Li₃N (s) + 3H₂0 (l) ► NH₃ (g) + 3LiOH (l)

                    1 mol                                                      3 mol

Convert moles to mass

Molar mass of LiOH = 24 g/mol

Molar mass of  Li₃N = 35 g/mol

So,

                   Li₃N (s)        +     3H₂0 (l) ► NH₃ (g) + 3LiOH (l)

                1 mol (35 g/mol)                                                   3 mol (24 g/mol)

                   35 g                                                                         72 g

So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of  LiOH

For this apply unity formula

                        35 g of  Li₃N ≅ 72 g of LiOH

                        X of  Li₃N ≅ 120 g of LiOH

By Doing cross multiplication

                  Mass of Li₃N = 35 g x 120 g / 72 g

                   mass of Li₃N =  58 g

120 g of LiOH will produce from 58 g of Li₃N

So,

Option D is correct =  58 g

there are two significant figures.