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 320.0 mg ND 3 ( g ) . The mass of deuterium, D , is 2.014 g/mol.

Question

contestada

Respuesta :

Аноним Аноним · Answer 1 · 2020-02-25T08:46:26+01:00

Answer: The mass of [tex]D_2O[/tex] needed is 0.96 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)

Given mass of [tex]ND_3[/tex] = 320.0 mg = 0.320 g (Conversion factor: 1 g = 1000 mg)

Molar mass of [tex]ND_3[/tex] = 20 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }ND_3=\frac{0.320g}{20g/mol}=0.016mol[/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, 0.016 moles of [tex]ND_3[/tex] will be produced from = [tex]\frac{3}{1}\times 0.016=0.048mol[/tex] of [tex]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 g/mol

Moles of [tex]D_2O[/tex] = 0.048 moles

Putting values in equation 1, we get:

[tex]0.048mol=\frac{\text{Mass of D_2O}}{20g/mol}\\\\\text{Mass of }D_2O}=(0.048mol\times 20g/mol)=0.96g[/tex]

Hence, the mass of [tex]D_2O[/tex] needed is 0.96 grams