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19greenlymoos
08.01.2020 •
Chemistry
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
Solved
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Ответ:
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
Ответ:
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Explanation:
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