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

Respuesta :

Answer:

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

Answer:

58g

Explanation:

just took the test! :D

have a miraculous day!! <3