Question

Ammonium nitrate explosively decomposes according to the following equation:

2 NH₄NO_3(s) --> 2 N_2(g)+O_2(g)+ 4 H₂O_(g)

Calculate the total volume of gas (at 123.8°C and 752 mmHg) produced by the complete decomposition of 1lb (455g) of ammonium nitrate (MW=80.05 g/mol).

Answer 1

Molar mass of NH4NO3 = 2*MM(N) + 4*MM(H) + 3*MM(O)

= 2*14.01 + 4*1.008 + 3*16.0

= 80.052 g/mol

mass of NH4NO3 = 455 g

mol of NH4NO3 = (mass)/(molar mass)

= 455/80.052

= 5.6838 mol

From balanced chemical reaction, we see that

when 2 mol of NH4NO3 reacts, 7 mol of gas is formed

mol of gas formed = (7/2)* moles of NH4NO3

= (7/2)*5.6838

= 19.89 mol

we have:

P = 752.0 mm Hg

= (752.0/760) atm

= 0.9895 atm

n = 19.89 mol

T = 123.8 oC

= (123.8+273) K

= 396.8 K

we have below equation to be used:

P * V = n*R*T

0.9895 atm * V = 19.89 mol* 0.08206 atm.L/mol.K * 396.8 K

V = 655 L

Answer: 655 L