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Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) How many moles of lithium are needed to produce 0.60 mol of Li3N when the reaction is carried out in the presence of excess nitrogen?

Answer :

Answer:

We need 1.80 moles of lithium

Explanation:

Step 1: Data given

Moles Li3N = 0.60 moles

Step 2: The balanced equation

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles Li

For 2 moles Li3N produced we need 6 moles Li and 1 mol N2

For 0.60 moles Li3N we need 3*0.60 = 1.80 moles Li

We need 1.80 moles of lithium

1.80 moles of lithium are needed to produce 0.60 mol of Li₃N when the reaction is carried out in the presence of excess nitrogen.

Given:

Moles of  Li₃N = 0.60 mol

Firstly write the balanced chemical equation:

6Li(s) + N₂(g) → 2Li₃N(s)

To find the number of moles of lithium:  

2 moles of  Li₃N are produced when 6 moles Li and 1 mol N₂ reacts.

⇒For 0.60 moles Li₃N, 3 x 0.60 = 1.80 moles Li are required.

1.80 moles of lithium are needed to produce 0.60 mol of Li₃N when the reaction is carried out in the presence of excess nitrogen.

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