Question

??2 (?) + ?2 (?) ⇋ ??3 (?) + ?2? (?) ?? = 1.23 × 10−3...

??2 (?) + ?2 (?) ⇋ ??3 (?) + ?2? (?) ?? = 1.23 × 10−3 at 534℃ ; 92.0 g ??2 (?) , 27,0 g ?2?(?) and 25,5 g ??3 (?) putting in a 750,0 mL balloon

What is the mole fraction of ?2? (?) in equilibrium?
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Answer #1

2 NO2 + 7 H2 2NH₂ + 4H₂O ko = [NH 3] ² [H2074 [NO₂] ² [H2]7 92 g Moles of NO2 - molar mass - 46 g/mol - = 279 = 2 mol moles oTotal 1.5 +1.5 moles = 2 + 20 68 + = 7.68 moles 8420 = moles of H₂O Total moles 1.5 7.68 [H2O = 0.19 mol = 0.2 mole fraction

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