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

Quizplus · Accepted Answer

To determine the limiting reactant, we need to compare the actual mole ratio of the reactants to the stoichiometric mole ratio.Actual mole ratio:125.0 g NH3 * (1 mol NH3 / 17.03 g NH3) = 7.34 mol NH3225.0 g O2 * (1 mol O2 / 32.00 g O2) = 7.03 mol O2Stoichiometric mole ratio:4 mol NH3 : 5 mol O2Comparing the actual mole ratio to the stoichiometric mole ratio, we see that O2 is the limiting reactant because it has the lower mole ratio.Therefore, the amount of NO produced will be limited by the amount of O2 available.Using the stoichiometry of the balanced equation, we can calculate the moles of NO produced:7.03 mol O2 * (4 mol NO / 5 mol O2) = 5.62 mol NOConverting moles of NO to grams:5.62 mol NO * (30.01 g NO / 1 mol NO) = 168.8 g NOTherefore, the correct answer is B) 168.8 grams.

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

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