Calculate how many grams of heavy water are required to produce 385.0 mg of ND3(

Question

Calculate how many grams of heavy water are required to produce 385.0 mg of ND3(g). The mass of deuterium, D, is 2.014 g/mol.

General Chemistr Donald McQuarrie Peter A. Rock Ethan Gallogly University Science Books presented by Sapling Learning Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li,N(s) +3H20(1) NH,(g)+3LiOH(aq) Heavy water is water with the isotope deuterium in place of ordinary hydrogen, and its formula is D2O. The above reaction can be used to produce heavy ammonia, NDs(g), according to the equation, Li,N(s)+3D ND,(g) +3 LiODiaq Calculate how many grams of heavy water are required to produce 385.0 mg of ND3(g). The mass of deuterium, D, is 2.014 g/mol. Number g D2O

Explanation / Answer

Li3N + 3 D2O -----------> ND3 (g) + 3 LiOD

3 mol 1 mol

3x 20.028 g 20.042 g

= 60.084 g

? 385 mg = 0.385 g

From the above equation, we can say that

To produce 20.042 g of ND3, 60.084 g of D2O is required.

So, To produce 0.385 g of ND3, ? g of D2O is required.

? = (0.385 g of ND3/ 20.042 g of ND3) x 60.084 g of D2O

= 1.154 g of D2O

Therefore,

To produce 385 mg of ND3, 1.154 g of D2O is required.

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