Question

6. If 0.500 mol NaN3 react, what mass in grams of nitrogen will result?

2 NaN3 → 2 Na + 3 N2
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Answer 1

From 0.500 mol of NaN3, 21.015 grams of nitrogen will be produced.

Step-by-step explanation:

According to the balanced equation:

2 NaN3 -> 2 Na + 3 N2

From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2. So, in order to find how many moles of N2 will be produced from 0.500 mol of NaN3, we can set up a proportion:

(3 mol N2) / (2 mol NaN3) = (x mol N2) / (0.500 mol NaN3)

Solving for x, we can find that x = 0.750 mol N2.

To convert moles to grams, we can use the molar mass of nitrogen, which is 28.02 g/mol. So, the mass of nitrogen produced would be:

0.750 mol N2 × 28.02 g/mol = 21.015 g

Therefore, the mass in grams of nitrogen produced from 0.500 mol of NaN3 is 21.015 grams.

Answer 2

21 g of N₂ are produced by the decomposition

Step-by-step explanation:

The reaction is: 2 NaN3 → 2 Na + 3 N2

2 moles of sodium nitride decompose in order to produce 2 moles of Na and 3 moles of nitrogen gas.

According to stoichiometry, ratio is 2:3. Therefore we say,

2 moles of nitride can produce 3 moles of N₂

Then, 0.5 moles of NaN₃ will produce (0.5 . 3) / 2 = 0.75 moles of N₂

We convert the moles to mass, to find the answer

0.75 mol . 28 g / 1 mol = 21 g