Question

`320` `g` of solid ammonium nitrite, `NH_4NO_2`, decomposes when heated according to the balanced equation: `NH_4NO_2 -> N_2+2H_2O`. What total volume of gases at `819` `K` is emitted by this reaction?

Answer 1

Each mole of `NH_4NO_2` produces `3` `mol` of gases total (`N_2` and `H_2O`), and we have `5` `mol`. So `3xx5 = 15` `mol` of gas at STP is `336` `L`. Converting to `819` `K` gives a volume of `1008` `L`.

Explanation:

The molar mass of `NH_4NO_2` is `64` `gmol^-1` (two `N` at `14`, four `H` at `1`, two `O` at `16`).

Find the number of moles of `NH_4NO_2`:

`n=m/M=320/64=5` `mol`

From the balanced equation, each mole of `NH_4NO_2` yields 1 mole of `N_2` and 2 moles of `H_2O` (which is definitely a gas at `819` `K`), for a total of 3 moles of gas.

For ideal gases (which we can treat these real gases as for our purposes), 1 mole of any gas at STP (`273` `K` and `1` `atm`) occupies `22.4` `L`.

That means we have `3xx5 = 15` `mol` of product gases, and they occupy a volume of `336` `L` at STP.

We need to change the temperature from `273` `K` to `819` `K`, and the pressure stays the same at `1` `atm`, so:

`V_1/T_1=V_2/T_2`

Rearranging:

`V_2=V_1T_2/T_1 = 336 xx 819/273 = 1008` `L`