Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
The crushed tablets would stop bubbling/fuzzing first because it has a smaller surface area which means that it would dissolve before the uncrushed tablets which has a larger surface area.
Chemical energy is converted to thermal energy.
Answer:
18.9 moles
Explanation:
We have the following data:
V = 50 L
P = 12.4 atm
T= 127°C + 273 = 400 K
R = 0.082 L.atm/K.mol (it is the gas constant)
We use the ideal gas equation to calculate the number of moles n of the gas:
PV = nRT
⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol
