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How many grams of magnesium nitride is required to produce 25.00 g of magnesium hydroxide?

Mg3N2 + H2O -> Mg(OH)2 + NH3

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Answer:

25 grams of Mg(OH)2 will be produced by 14.424 gram of Mg3N2

Step-by-step explanation:

The balanced equation is

Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3

Molecular weight of magnesium nitride = 100.9494 g/mol

Molecular weight of magnesium hydroxide = 58.3197 g/mol

one mole of Mg3N2 produces three moles of 3Mg(OH)2

100.9494 g/mol of Mg3N2 produces 3* 58.3197 g/mol of Mg(OH)2

1 gram of Mg3N2 produces


(3* 58.3197)/(100.9494 ) \\1.733grams of Mg(OH)2

Or 1.733 grams of Mg(OH)2 will be produced by 1 gram of Mg3N2

25 grams of Mg(OH)2 will be produced by 14.424 gram of Mg3N2

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