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

Question

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

Quizplus · Accepted Answer

This is a limiting reactant problem where we need to determine which reactant gets completely used up first, thus limiting the amount of product that can be formed. To solve, we first need to convert the given masses of NH₃ and O₂ to moles using their respective molar masses. 125.0 g NH₃ x (1 mol / 332.46 g) = 0.3750 mol NH₃ 225.0 g O₂ x (1 mol / 284.48 g) = 0.7914 mol O₂ Next, we need to determine the mole ratio of the reactants using the coefficients in the balanced chemical equation. 4 moles NH₃ : 5 moles O₂ Using this ratio, we can determine how many moles of NH₃ will react with the given amount of O₂. 0.7914 mol O₂ x (4 mol NH₃ / 5 mol O₂) = 0.6331 mol NH₃ consumed Since there was less NH₃ present than the amount consumed, it is the limiting reactant. We can now use the mole ratio from the balanced chemical equation to determine the amount of O₂ that reacts with the limiting amount of NH₃. 0.3750 mol NH₃ x (5 mol O₂ / 4 mol NH₃) = 0.469 mol O₂ consumed Finally, we can subtract the amount of NH₃ consumed from the initial amount to determine the amount remaining after the reaction. 0.3750 mol NH₃ - 0.3750 mol NH₃ = 0 mol NH₃ Therefore, the answer is B) 29.4 grams of NH₃ will be present after a complete reaction.

Consider the Reaction: 4 NH3 + 5 O2 → 6

Consider the Reaction: 4 NH₃ + 5 O₂ → 6