Automotive air bags inflate when sodium azide, NaN3, rapidly decomposed into N2 and Na by the following reaction: 2 NaN3 --> 2 Na + 3 N2. In order to fill an air bag, 75.0 grams of N2 must be formed.

Question

Question: How many grams of NaN3 must react to form 75.0 grams of N2.

Answer 1

first convert 75.0 g to Moles by dividing by 28g/mol. = 2.68 moles N₂

Based on the balanced equation reactants to product ratio are 2:3 for Azide: Nitrogen gas. This means for every 3 moles of Nitrogen gas produces, 2 moles of Sodium Azide are needed.

This is a dimensional analysis question by using the fence method.

Start with moles calculated from above.

2.68 Moles N₂ *(2 moles NaN₃/3 moles N₂)=1.79mol NaN₃(65g/mol)= 116.35 g or with sig. figs 116g NaN₃

Answer 2

For this problem, we will need to work our way backwards from 75.0 g N₂, to moles N₂, to moles NaN₃, and finally to grams NaN₃.

We can convert mass of N₂ to moles of N₂ using the molar mass of nitrogen gas, We can determine the molar mass using the atomic mass of nitrogen from the periodic table:

molar mass of N₂: 2 * atomic mass nitrogen = 2 * (14.01 g / mol) = 28.02 g / mol

Now we convert our desired mass of nitrogen gas to moles:

75.0 g (1 mol / 28.02 g) = 75.0 / 28.02 g ( mol / g) = 2.68 moles N₂

Next we need to convert moles of nitrogen gas to moles of sodium azide. Let's take a look at our balanced reaction:

2 NaN₃ → 2 Na + 3 N₂

We can see that for every 2 moles of sodium azide, we will form 3 moles of nitrogen gas.

2.68 mol N₂ * (2 mol NaN₃ / 3 mol N₂) = 2.68 * 2 / 3 mol N₂ * (mol NaN₃ / mol N₂) = 1.79 mol NaN₃

Finally, we determine the molar mass of sodium azide and convert from moles to mass:

molar mass sodium azide: 1 ( atomic mass sodium) + 3 (atomic mass nitrogen) = (22.99 + 3*14.01) g / mol = 65.02 g / mol NaN₃

1.79 mol NaN₃ (65.02 g / mol NaN₃) = 1.79 * 65.02 mol NaN₃ (g / mol NaN₃) = 116 g NaN₃