Question

Ammonium nitrate decomposes explosively upon heating according to the following balanced equation:

2NH4NO3(s)→2N2(g)+O2(g)+4H2O(g)
Calculate the total volume of gas (at 120 ∘C and 766 mmHg ) produced by the complete decomposition of 1.44 kg of ammonium nitrate.

Answer 1

PV=nRT0.988×V=76.56×0.0821×394V=2506.6 L

Step-by-step explanation:

• amount of ammonium nitrate present is m = 1.75 kg = 1750g.

• corruption of ammonium nitrate upon heating

Reaction 2NH4NO3( s) ⟶ 2N2( g) O2( g) 4H2O( g)

Molar Mass M 80 g/mol

StoichiometricCoefficient( n) 2 2 1 4

Stoichiometric Mass m = ( n × M) ( 2 × 80) g = 160 g

From the stoichiometric mass we have 160 g of ammonium nitrate produces( 2 1 4) intelligencers ie 7 moles of gas.

thus we have number of intelligencers of feasts evolved from 1750 g of ammonium nitrate equal to

n = 7/160 × 1750 moles= 76.56 moles

• Pressure of the gases are

• P = 751 mmHg = 0.988 atm

Note 1 atm = 760 mmHg.

• Temperature of the gases are

T = 121oC = 394 K

Let the volume of feasts produced be V.

From the ideal gas equation we've

PV = nRT

0.988 × V = 76.56 ×0.0821 × 394

V = 2506.6 L