What is the FORMULA for the limiting reagent? What amount of the excess reagent

Question

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?

For the following reaction, 15.4 grams of magnesium nitride are allowed to react with 12.8 grams of water. magnesium nitride (s) + water (l) rightarrow magnesium hydroxide (aq) + ammonia (aq) What is the maximum amount of magnesium hydroxide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete?

Explanation / Answer

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

Mg3N2 moles = 15.4/100.9 = 0.1526

12.8 gm water = 12.8/18 = 0.71

H2O is limiting reagent

Mg(OH)2 moles = 0.71 x 3/6 = 0.355 , Mg(OH) 2 max = 0.355 x 58.32 = 20.7 gm

excess Mg3N2 = ( 0.1526 ) -( 0.71/6) = 0.0342 moles

excess Mg3N2 = 0.0342 x 100.9 = 3.457 gm

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