Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation Li 3 N ( s ) + 3 H 2 O ( l ) ⟶ NH 3 ( g ) + 3 LiOH ( aq ) Heavy water is water with the isotope deuterium in place of ordinary hydrogen, and its formula is D 2 O . The same reaction can be used to produce heavy ammonia, ND 3 ( g ) , according to the equation Li 3 N ( s ) + 3 D 2 O ( l ) ⟶ ND 3 ( g ) + 3 LiOD ( aq ) Calculate how many grams of heavy water are required to produce 455.0 mg ND 3 ( g ) . The mass of deuterium, D , is 2.014 g/mol.

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Аноним Аноним · Answer 1 · 2020-03-09T05:24:09+01:00

Answer: The mass of [tex]D_2O[/tex] needed is 1.36 grams

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)

For [tex]ND_3[/tex] :

Given mass of [tex]ND_3[/tex] = 455.0 mg = 0.455 g (Conversion factor: 1 g = 1000 mg)

Molar mass of [tex]ND_3[/tex] = 20.049 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }ND_3=\frac{0.455g}{20.049g/mol}=0.0227mol[/tex]

For the given chemical equation:

[tex]Li_3N(s)+3D_2O(l)\rightarrow ND_3(g)+3LiOD(aq.)[/tex]

By Stoichiometry of the reaction:

1 mole of [tex]ND_3[/tex] is produced from 3 moles of [tex]D_2O[/tex]

So, 0.0227 moles of [tex]ND_3[/tex] will be produced from = [tex]\frac{3}{1}\times 0.0227=0.0681mol\text{ of }D_2O[/tex]

Now, calculating the mass of [tex]D_2O[/tex] from equation 1, we get:

Molar mass of [tex]D_2O[/tex] = 20.027 g/mol

Moles of [tex]D_2O[/tex] = 0.0681 moles

Putting values in equation 1, we get:

[tex]0.0681mol=\frac{\text{Mass of }D_2O}{20.027g/mol}\\\\\text{Mass of }D_2O}=(0.0681mol\times 20.027g/mol)=1.36g[/tex]

Hence, the mass of [tex]D_2O[/tex] needed is 1.36 grams