Question

# A 10.00 g of sodium metals reacts with 2.50-l of nitrogen gas at 0.976 atm and...

A 10.00 g of sodium metals reacts with 2.50-l of nitrogen gas at 0.976 atm and 28 degree C to produce sodium nitride. Write a balanced chemical equation for this reaction . Do not include phases. part b : How many grams of sodium nitride are produced in this reaction

The balanced equation is

6 Na + N2 ------> 2 Na3N

Number of moles of Na = 10.00 g / 22.9897 g/mol = 0.435 mole

PV = nRT

T = 28 + 273 = 301 K

0.976 * 2.50 = n * 0.0821 * 301

2.44 = n * 24.7

n = 2.44 / 24.7 = 0.0988 mole

Therefore, number of moles of N2 = 0.0988 mole

From the balanced equation we can say that

6 mole of Na requires 1 mole of N2 so

0.435 mole of Na will require

= 0.435 mole of Na *(1 mole of N2 / 6 mole of Na)

= 0.0725 mole of N2

But we have 0.0988 mole of N2 which is in excess so N2 is excess reactant and Na is limiting reactant

From the balanced equation we can say that

6 mole of Na produces 2 mole of Na3N so

0.435 mole of Na will produce

= 0.435 mole of Na *(2 mole of Na3N / 6 mole of Na)

= 0.145 mole of Na3N

mass of 1 mole of Na3N = 82.98 g

so the mass of 0.145 mole of Na3N = 12.0 g

Therefore, the mass of Na3N produced would be 12.0 g

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