## If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, which reactant

Question

If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, which reactant would be the limiting reactant? (4 points) How much NaN3 would actually be produced?

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2 weeks 2021-08-31T12:05:33+00:00 1 Answers 0 views 0

N₂

Explanation:

Sodium is a larger molecule with a much higher molecular weight. However, 20g of N₂ would be a smaller amount of molecules than 20g of sodium due to how there are multiple nitrogen molecules.

2/3 * given mass of N₂ = mass of N₃

N₃ (Azide ion) given mass = 20g

Na = 20g

Masses of chemicals are equal

Na = 22.990g/mol

20g/22.990g/mol = 0.8699mol of Na

N₃ = 20g

N₃ = N g/mol x 3

N = 14.007 g/mol

14.007 x 3 = 42.021 g/mol

N₃ = 42.021g/mol

20g/42.021g/mol = 0.4759 mol of N₃

Notice how there are fewer moles of the Azide ion than the Sodium.

0.4759 moles of NaN₃ is produced

combine molecular weights:

42.021 + 22.990 = 65.011 g/mol

multiply by amount of moles of the limiting reactant:

0.4759 mol *65.011 = 30.942 g

Also, here is the balanced equation:

3N₂ + 2Na = 2NaN₃

The results are the same as the balanced equation.