Question

[Help, Help, Help! Stoichiometry and Gases, please explain]

The following reaction takes place at STP. 4.40 L of nitrogen gas are formed. How many grams of ammonium nitrite decomposed?

2 NH4NO2 (aq) ---> 2 N2 (g) + 4 H2O (l)​

Answer 1

At STP, it is at 273 K and 1.00 atm.
Use PV=nRT to and solve for n (number of mol of N2)
1.00 atm • 4.40L / 0.08206 Latm/molK • 273 K = 0.196 mol N2
There are two mol of N2 for every two mol of NH4NO2 (the number of mol of both are equal)
Find the molar mass of NH4NO2 = 64.1 g
Multiply the number of mol by the molar mass
0.196 mol • 64.1 = 12.6 g NH4NO2

Answer 2

The mass of ammonium nitrite decomposed is 12.54 grams.

Step-by-step explanation:

Pressure of the nitrogen gas = P = 1 atm

Volume of the nitrogen gas = V = 4.40 L

Temperature of the nitrogen gas = T = 273.15 K

Moles of nitrogen gas = n

PV = nRT (ideal gas equation)

`n=(PV)/(RT)=(1 atm* 4.40 L)/(0.0821 atm L/mol K* 273.15 K)=0.196 mol`

`2 NH_4NO_2 (aq)\rightarrow 2N_2 (g) + 4 H_2O (l) ​`

According to reaction , 2 mole of nitrogen is obtained from 2 mole of ammonium nitrite.

Then 0.196 moles of nitrogen will will be obtained from:

`(2)/(2)* 0.196 mol=0.196 mol` ammonium nitrite.

Mass of 0.196 moles of ammonium nitrite:

= 0.196 mol × 64 g/mol = 12.54 g

The mass of ammonium nitrite decomposed is 12.54 grams.