Question

Calculate the total volume of gas (at 134 ∘C and 748 mmHg ) produced by the complete decomposition of 1.45 kg of ammonium nitrate.

Answer 1

Solution.

. Decomposition of Nhy N03 (ammoniom Nitrate) is given by 2NH4NO3 (5) ► 2N269)+ O2(g) + 4 H2009) There are 2+1+4 = 1 moles of gas poroduced Molus mass of NHYNO3 = 80.043 g|mol Given Weight = 1.45 kg Converting Kg unto g. 1 kg = 1000g 1.45 kg = 145x1000g = 1.45X1087 Colculate number of moles of goses produced (un Total). Molas mass of Ny Nog gibos Contains i mole of NH4NO3 ie 80.0439 of why NOg have I mole of NHYNCZ 1 → potoue mole LUSX1089" 145X10% mole w : → 18.11 mole of N HY NOz. Means 145x10°g of WHY NOg have 18.11 mole of WHY NO3 80.043

Calculating number of moles of gas — ngas. from střechiometry it is clear that. 2 moles of NHy Noz gives - 7 moles of gas. I mole " " 7 moles of gas. 18.11 moles "" → ZX18.11 " Maas → 63.385 moks of gas. Accosding to Ideal gas law PV = nRT Py Pressure vt volume T= 134°C = (134+273) K n- Number of moles R Gas constant p=748 mmHg Ry 0.0821 Latom k moet T= Temperature I atm = 760 mmHg] 748 mmHg → 748 nasty x 1 atm = 0.9842 atm o d 760 namita Putting the values eivý expression - Vyas = ngas RT ngas RT = 63.385 mootx 0.082) Latin x 407K K baot 0.9842 chan 1 Vzas = 2:15X103 Answer = 407K ор