Automotive airbags inflate when sodium azide NaN3 decomposes. In preparation of the

1 answer:

8 0

63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction

The balanced reaction equation is;

2 Na + 3 N2--> 2 NaN3

Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na

Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2

We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.

For Na

2 moles of Na yields 2 moles of NaN3

2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).

For N2

3 moles of N2 yields 2 moles of NaN3

1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3

So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.

Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3

Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction

Learn more: brainly.com/question/9743981

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