Question

What volume in (L) of N2 would be produced if 143 g of NaN3 completely reacted at STP?

Answer 1

INFORMATION:

We know that:

- An automobile airbag inflates when NaN3 is converted to Na and N2 gas according to the equation,

2NaN3 → 2 Na + 3 N2.

And we must find the volume in (L) of N2 if 143 g of NaN3 completely reacted at STP

STEP BY STEP EXPLANATION:

To find it, we need to work with the conditions as an ideal gas.

The ideal gas equation is

`PV=nRT`

1. We must find n:

To find n, we must first divide the 143 g of NaN3 by the molar mass

`(143g)/(65(g)/(mol))=2.2mol`

So, we have 2.2 mol of NaN3

Now to find n, we must multiply the number of mol of NaN3 by the relation between NaN3 and N2

`2.2mol\text{ }NaN_3*\frac{3mol\text{ }N_2}{2mol\text{ }NaN_3}=3.3mol\text{ }N_2`

Then, n = 3.3 mol

2. using STP conditions:

- T = 273K

- P = 760 torr

And using R = 62.36 (L*torr)/(K*mol)

Finally, replacing in the ideal gas equation solver for V

`V=(nRT)/(P)=(3.3mol*62.36(L\cdot torr)/(K\cdot mol)*273K)/(760torr)=73.92L`

So, 73.92 L of N2 would be produced if 143 g of NaN3 completely reacted at STP.

ANSWER:

73.92 L