N2H4(l) + O2(g) --> N2(g) + 2H2O (l); ΔH= -285.8 kJ How many kJ of heat...

N₂H₄(l) + O₂(g) --> N₂(g) + 2H₂O (l); ΔH= -285.8 kJ

How many kJ of heat will be released when 35.7 g water is generated?

FW: N = 14; H = 1; O = 16.

Expert Solution

Given: Heat released for the reaction ∆H = -285.8 kJ
Mass of water, H₂O generated = 35.7 g

N₂H₄ (l) + O₂ (g) → N₂ (g) + 2 H₂O (l)

Molar mass of water, H₂O = 2 x Mass of H + Mass of O
= 2 x 1 + 16 = 18g
Number of moles of water, H₂O generated = Given mass of water, H₂O / Molar mass of water, H₂O
= 35.7 / 18 = 1.98 mol

As is clear from the chemical equation and the given data, for 2 mol of water, H₂O generated 285.8 kJ of heat is released
⇒ For 1 mol of water, H₂O generated (285.8 / 2) kJ of heat is released
⇒ For 1.98 mol of water, H₂O generated (285.8 / 2) x 1.98 kJ of heat is released
⇒ For 1.98 mol of water, H₂O generated 283.42 kJ of heat is released

Therefore heat released, ∆H = -283.42 kJ

queen_honey_blossom answered 2 years ago

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