Question

What is the total volume of products obtained when 128 g NH4NO2 decomposes at 819 o C and 2.00 atm

NH4NO2 (s) → N2 (g) + 2H2O (g)

Answer 1

NH₄NO₂ (s) → N₂ (g) + 2H₂O (g)

1 mole of NH₄NO₂ will produce 1 mole of N₂ and 2 moles of water.

Hence every 1 mole of NH₄NO₂ will result in 3 moles of gas

In the given condition,

Mass of NH₄NO₂ = 128 g

number of moles NH₄NO₂= mass/molar mass

Molar mass of NH₄NO₂ = 64 g/mol

∴Number moles of NH₄NO₂ required = 128 g / 64 g/mol = 2 moles

Therefore, the number of moles of gas produced by NH₄NO₂ = 3 * 2 = 6 moles

From the ideal gas equation,

PV = nRT

n = 6 moles

P = 2 atm (given)

T = 819°C = 273 +819 K (given)

   = 1092 K

Gas constant, R = 0.08205 L atm mol^-1 K^-1

                             Therefore, 2 atm x V = 6 moles x 0.08205 L atm mol^-1 K^-1 x 1092 K

V = (6 moles x 0.08205 L atm mol^-1 K^-1 x 1092 K) / 2 atm

  V = 268.7958 L

Hence the volume of the products obtained is 268.80 L