Question

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

Answer 1

58g

Step-by-step explanation:

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Answer 2

Option D is correct = 58 g

Step-by-step 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