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How many grams of Li3N can be formed from 1.00 moles of Li? Assume an excess of nitrogen. The molar mass of lithium nitride is 34.83 g/mol. 6 Li (s) + N₂ (g) → 2 Li3N (s)

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Final answer:

From 1.00 moles of Li, considering an excess of nitrogen, 11.61 grams of Li3N can be formed using its molar mass of 34.83 g/mol based on the stoichiometric ratio in the balanced reaction.

Step-by-step explanation:

To calculate how many grams of Li3N can be formed from 1.00 moles of Li, given an excess of nitrogen and the molar mass of lithium nitride being 34.83 g/mol, we can use the stoichiometry of the balanced chemical reaction: 6 Li (s) + N₂ (g) → 2 Li₃N (s).

According to the stoichiometry, 6 moles of Li yield 2 moles of Li3N. Therefore, 1 mole of Li will yield 1/3 mole of Li3N. Calculating the mass of 1/3 mole of Li3N:

Mass = moles × molar mass = ⅓ moles × 34.83 g/mol = 11.61 g of Li3N

Thus, 11.61 grams of Li3N can be formed from 1.00 moles of Li, assuming there is an excess of nitrogen.

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