Question

For the reaction:

6Li(s)+N2(g)→2Li3N(s)
Determine:
the mass of N2 needed to react with 0.536 moles of Li.
the number of moles of Li required to make 46.4 g of Li3N.
the mass in grams of Li3N produced from 3.65 g Li.
the number of moles of lithium needed to react with 7.00 grams of N2.

Answer 1

To determine the required masses and mole quantities in the reaction of lithium and nitrogen, stoichiometric calculations based on the balanced chemical equation and molar masses are used.

Step-by-step explanation:

For the reaction 6Li(s) + N2(g) → 2Li3N(s), we can apply stoichiometric principles to solve each of the student's questions.

1. The mass of N2 needed to react with 0.536 moles of Li can be calculated using the molar ratio of Li to N2 from the balanced equation, which is 6:1. Since the molar mass of N2 is approximately 28.02 g/mol, we first determine the number of moles of N2 required for 0.536 moles of Li, then multiply by the molar mass of N2.
2. To determine the number of moles of Li required to make 46.4 g of Li3N, we use the molar mass of Li3N to find the moles of Li3N and then apply the molar ratio from the balanced equation to find the moles of Li.
3. For the mass in grams of Li3N produced from 3.65 g Li, we convert the mass of Li to moles, use the molar ratio to find moles of Li3N produced, and convert that to grams using the molar mass of Li3N.
4. The number of moles of lithium needed to react with 7.00 grams of N2 is found by converting the mass of N2 to moles and then using the balanced equation to find the corresponding moles of Li.

Answer 2

Explanation:

`6Li(s)+N_2(g)\rightarrow 2Li_3N(s)`

1. Mass of
`N_2` needed to react with 0.536 moles of Li.

According to reaction, 6 moles of Li reacts with 1 mol of
`N_2`.

Then 0.536 moles of Li will react with:

`(1)/(6)* 0.536` moles of
`N_2` that is 0.0893 moles.

Mass of
`N_2[tex] gas needed:</strong></p><p><strong>[tex]=28 g/mol* 0.0893 mol=2.5004 g`

2.The number of moles of Li required to make 46.4 g of
`Li_3N`

Moles of
`Li_3N=(46.4 g)/(35 g/mol)=1.3257 mol`

According to reaction the 2 moles of
`Li_3N` are produced from 6 moles of Li.

Then 1.3257 moles of
`Li_3N` will produced from:

`(6)/(2)* 1.3257=3.9771 moles`

3.9771 moles of lithium will needed.

3. The mass in grams of
`Li_3N` produced from 3.65 g Li.

Moles of Li
`=(3.65 g)/(7 g/mol)=0.5214 moles`

According to reaction, 6 moles of Li gives 2 moles of
`Li_3N`

Then 0.5214 moles of Li will give
`(2)/(6)* 0.5214` that is 0.1738 moles of
`Li_3N`.

Mass of
`Li_3N=0.1738 mole* 35 g/mol=6.083 g`

6.083 grams of
`Li_3N` will be produced.

4. The number of moles of lithium needed to react with 7.00 grams of
`N_2`.

Moles of
`N_2=(7.00 g)/(28 g/mol)=0.25 moles`

1 mol of
`N_2` reacts with 6 mol of Li

Then, 0.25 moles of
`N_2` will ftreact with :

`6* 0.25 moles=1.5 moles` of lithium

1.5 moles of Li will be needed.