For the reaction: 4NH₃ + 5O₂ ® 6H₂O + 4NO
If 125.0 g of NH₃ is reacted with 225.0 g of O₂, how many grams of H₂O will be present after a complete reaction?

Question

For the reaction: 4NH₃ + 5O₂ ® 6H₂O + 4NO If 125.0 g of NH₃ is reacted with 225.0 g of O₂, how many grams of H₂O will be present after a complete reaction?

Quizplus · Accepted Answer

The balanced equation shows that 4 moles of NH3 react with 5 moles of O2 to produce 6 moles of H2O. The molar mass of NH3 is 17.03 g/mol, and the molar mass of O2 is 32.00 g/mol. Therefore, 125.0 g of NH3 is equal to 125.0 g / 17.03 g/mol = 7.34 mol of NH3, and 225.0 g of O2 is equal to 225.0 g / 32.00 g/mol = 7.03 mol of O2. Since there are fewer moles of O2 than NH3, O2 is the limiting reactant. Therefore, the maximum number of moles of H2O that can be produced is 7.03 mol. The molar mass of H2O is 18.02 g/mol, so 7.03 mol of H2O is equal to 7.03 mol * 18.02 g/mol = 126.6 g of H2O.

For the Reaction: 4NH3 + 5O2 ® 6H2O + 4NO

For the Reaction: 4NH₃ + 5O₂ ® 6H₂O + 4NO