Calculate the total volume of gas (at 119 ? c and 731 mmhg produced by the complete decomposition of 1.71 kg of ammonium nitrate.

The reaction is:

NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)

This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.

Now find the number of moles in 1.71 kg of NH4 (NO3)

Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol

# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)

We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol

Now use the equation for ideal gases to fin the volume

pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =

V = 2143.01 liters

No is the answer to the question