If 0.500 mol NaN3 react, what mass in grams of nitrogen would result?

looks like 3*.5 mol of N, or 1.5*.5 N2

convert that to grams.

To determine the mass of nitrogen produced when 0.500 mol of NaN3 reacts, we first need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between NaN3 and the formation of nitrogen is as follows:

2 NaN3 → 2 Na + 3 N2

From the balanced equation, we can determine that for every 2 moles of NaN3 that react, 3 moles of N2 are produced. This means that:

2 mol NaN3 → 3 mol N2

Now, we can set up a proportion using the given amount of NaN3:

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

Cross-multiplying and solving for x, we get:

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

x = 0.750 mol N2

Finally, to convert moles of N2 to grams, we need to use the molar mass of nitrogen, which is approximately 28.0134 g/mol. Multiplying the number of moles of N2 by the molar mass, we get:

Mass of nitrogen = (0.750 mol N2) * (28.0134 g/mol)

Mass of nitrogen = 21.01 g

Therefore, 0.500 mol of NaN3 will result in 21.01 grams of nitrogen.

To determine the mass of nitrogen produced when 0.500 mol of NaN3 reacts, we need to use the balanced chemical equation for the reaction:

2 NaN3(s) -> 3 N2(g) + 2 Na(s)

From the balanced equation, we can see that 2 moles of NaN3 produce 3 moles of N2.

Now, we can set up a proportion to find the moles of nitrogen produced:

2 moles NaN3 / 3 moles N2 = 0.500 moles NaN3 / x

Cross-multiplying the equation:

2 * x = 3 * 0.500

2x = 1.500

x = 1.500 / 2

x = 0.750 moles of N2

Lastly, to find the mass of nitrogen in grams, we can use the molar mass of nitrogen, which is approximately 28 grams per mole:

Mass of nitrogen = moles of N2 * molar mass of N2

Mass of nitrogen = 0.750 moles * 28 grams/mole

Mass of nitrogen = 21 grams

Therefore, when 0.500 mol of NaN3 reacts, the mass of nitrogen produced is 21 grams.