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For the reaction

6Li+ N2 —> 2Li3N

What us the maximum amount of Li3B which could be formed from 7.87 mol of Li and 4.37 mol of N2?

Answer in units of mol.

Respuesta :

Neetoo

Answer:

Mass = 90.6 g

2.6 mol

Explanation:

Given data:

Number of moles of Li = 7.87 mol

Number of moles of N = 4.37 mol

Mass of Li₃N = ?

Solution

Chemical equation:

6Li + N₂   →   2Li₃N

Now we will compare the moles of nitrogen and lithium with Li₃N.

                            N₂              :          Li₃N

                             1               :             2

                           4.37           :          2×4.37 =8.74

                            Li               :               Li₃N

                             6               :                2

                            7.87           :             2/6×7.87 = 2.6 mol

Mass of Li₃N:

Mass = number of moles × molar mass

Mass = 2.6 mol × 34.83 g/mol

Mass = 90.6 g

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