For the following reaction, 5.57 grams of magnesium nitride are mixed with excess water. The reaction yields 6.89grams of magnesium hydroxide.Mg3N2 (s) + 6H2O (l) 3Mg(OH)2 (aq) + 2NH3 (aq)(1) What is the theoretical yield of magnesium hydroxide?grams(2) What is the percent yield for this reaction?

Given data, Mass of magnesium nitride (Mg3N2) = 5.57 gram Mg3N2(s) + 6H2O(l) <====>…

Answered: For the following reaction, 5.57 grams…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Question

For the following reaction, 5.57 grams of magnesium nitride are mixed with excess water. The reaction yields 6.89grams of magnesium hydroxide.

Mg₃N₂ (s) + 6H₂O (l) 3Mg(OH)₂ (aq) + 2NH₃ (aq)

(1) What is the theoretical yield of magnesium hydroxide?	grams
(2) What is the percent yield for this reaction?

Expert Solution

Step 1

Given data,

Mass of magnesium nitride (Mg₃N₂) = 5.57 gram

Mg₃N₂(s) + 6H₂O(l) <====> 3Mg(OH)₂(aq) + 2NH₃(aq)

Percent yield = ( Actual yield/ Theoretical yield) ×100 %

Actual yield of magnesium hydroxide Mg(OH)₂ = 6.89 gram

Number of moles of Mg₃N₂ = Mass of Mg₃N₂ / molar mass of Mg₃N₂

Number of moles of Mg₃N₂ = 5.57 g/100.94 g.mol^-1

Number of moles of Mg₃N₂= 0.05518 mol

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