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
First choice: PhH4 has a lower boiling point
Second: SiF4 higher
Third: CBr4 higher
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