if excess NH3 is reacted with 0.438 mol of O2, how many moles of H2O will be produced?
4NH3+7O2-->4NO2+6H2O
0.438 mols O2 x (6 mols H2O/7 mols O2) = 0.438 x (6/7) = ?
To determine the number of moles of H2O produced when excess NH3 (ammonia) reacts with 0.438 mol of O2 (oxygen), you need to use the balanced chemical equation provided:
4NH3 + 7O2 --> 4NO2 + 6H2O
From the equation, you can see that 7 moles of O2 react to produce 6 moles of H2O. Thus, the mole ratio of O2 to H2O is 7:6.
To find the number of moles of H2O produced, you can use the following calculations:
0.438 mol O2 x (6 mol H2O / 7 mol O2) = 0.375 moles H2O
Therefore, when excess NH3 reacts with 0.438 mol of O2, approximately 0.375 moles of H2O will be produced.