Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) How many moles of lit hium are needed to produce 0.60 mol of Li3N when the reaction is carried out in the presence of excess nitrogen?
2 answers:
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
Answer:
12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N
Explanation:
The reaction is:
6Li(s) + N₂(g) → 2Li₃N(s)
If nitrogen is in excess, the lithium is the limiting reactant.
Ratio is 2:6
2 moles of nitride were produced by 6 moles of Li
Then, 0.6 moles of nitride were produced by (0.6 .6)/ 2 = 1.8 moles of Li
Let's convert the moles to mass → 1.8 mol . 6.94 g/ 1mol = 12.5 g of Li
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