Question

Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li (s) + N2 (g) → 2Li3N (s) In a particular experiment, 4.00-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Answer 1

Answer: 4.02

Step-by-step explanation:

-- 4.00 g Li x 1 mol Li x 2 mol LiN x 20.948 g LiN = 4.02 g LiN

-------------- ---------------- ---------------------

6.941 g Li 6 mol Li 1 mol LiN

-- 4.00 g N2 x 1 mol N2 x 2 mol LiN x 20.948 g LiN = 5.98 g LiN

-------------- ---------------- ---------------------

28.014 g N2 1 mol N2 1 mol LiN

Answer 2

Answer: 6.71 g

Step-by-step explanation:
`6Li(s)+N_2(g)\rightarrow 2Li_3N`

`{\text{no of moles}}=\frac{\text{Given mass}}{\text{Molar mass}}`

`{\text {moles of lithium}}=(4g)/(6.914g/mol)=0.578moles`

`\text{moles of nitrogen}=(4g)/(28g/mol)=0.143moles`

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus lithium is the limiting reagent and nitrogen is the excess reagent.

As can be seen from the balanced chemical equation, 6 moles of lithium reacts with 1 mole of nitrogen to give 2 moles of lithium nitride.

Thus 0.578 moles of lithium react with 0.096 moles of nitrogen.

6 moles of lithium give = 2 moles of lithium nitride

Thus 0.578 moles of lithium give=
`(2)/(6)* {0.578}=0.19moles` of lithium nitride.

Mass of lithium nitride
`Li_3N={\text {no of moles}}* {\text {Molecular mass}}`

Mass of lithium nitride
`Li_3N`=
`{0.192moles}* {34.83g/mol}=6.71g`