# How many grams of li3n can be formed from 1.75 moles of li?

How many grams of li3n can be formed from 1.75 moles of li? assume an excess of nitrogen. 6 li(s) + n2(g) â 2 li3n(s) how many grams of li3n can be formed from 1.75 moles of li? assume an excess of nitrogen. 6 li(s) + n2(g) â 2 li3n(s) 18.3 g li3n 61.0 g li3n 58.3 g li3n 20.3 g li3n 15.1 g li3n?

20.3 g First, determine how many moles of Li3N you can produce from 1.75 moles of Li. Since each molecule of Li3N requires 3 atoms of Li, that means that from 1.75 moles of Li, you can only produce 1.75 / 3 = 0.5833 moles of Li3N. Now compute the molar mass of Li3N Atomic weight of Lithium = 6.941 Atomic weight of Nitrogen = 14.0067 Molar mass of Li3N = 3 * 6.941 + 14.0067 = 34.8297 g/mol Now multiply the molar mass by the number of moles 34.8297 * 0.5833 = 20.316 g Of the available 5 choices, only 20.3 g is correct.
The balance chemical reaction is:

6 Li(s) + N2(g) → 2 Li3N(s)

We use the amount of the Lithium reactant and the relations of the substances from the reaction to calculate the amount of product produced. We do as follows:

1.75 mol Li ( 2 mol Li3N / 6 mol Li ) (34.83 g / mol) = 20.32 g Li

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