if excess NH3 is reacted with 0.438 mol of O2, how many moles of H2O will be produced?
4NH3+7O2-->4NO2+6H2O
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0.438 mols O2 x (6 mols H2O/7 mols O2) = 0.438 x (6/7) = ?
4NH3+7O2-->4NO2+6H2O
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1 answer
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