42.6 g Cu are combined with 84.0 g of HNO₃ according to the reaction: 3 Cu + 8 HNO₃ → 3 Cu(NO₃) ₂ + 2 NO + 4 H₂O.
Which reagent is limiting and how many grams of Cu(NO₃) ₂ are produced?

Question

42.6 g Cu are combined with 84.0 g of HNO₃ according to the reaction: 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O. Which reagent is limiting and how many grams of Cu(NO₃)₂ are produced?

Quizplus · Accepted Answer

First, we need to determine which reagent is limiting, so we can calculate the maximum amount of product produced. We can do this by calculating the amount of product that would be formed if all of the Cu reacted and the amount of product that would be formed if all of the HNO₃ reacted, and then comparing the two values. To calculate the amount of product formed if all of the Cu reacted, we need to convert the mass of Cu to moles, and then use stoichiometry to calculate the moles of Cu(NO₃)₂ that would be formed: 42.6 g Cu / 63.55 g/mol Cu = 0.6707 mol Cu 0.6707 mol Cu × (3 mol Cu(NO₃)₂ / 3 mol Cu) = 0.6707 mol Cu(NO₃)₂ To calculate the amount of product formed if all of the HNO₃ reacted, we need to convert the mass of HNO₃ to moles, and then use stoichiometry to calculate the moles of Cu(NO₃)₂ that would be formed: 84.0 g HNO₃ / 149.09 g/mol HNO₃ = 0.5631 mol HNO₃ 0.5631 mol HNO₃ × (3 mol Cu(NO₃)₂ / 8 mol HNO₃) = 0.2099 mol Cu(NO₃)₂ Based on these calculations, we can see that the amount of product formed by the complete reaction of Cu is higher than the amount formed by the complete reaction of HNO₃, so HNO₃ is the limiting reagent. Next, we need to calculate the actual amount of product formed, using stoichiometry and the amount of limiting reagent: 84.0 g HNO₃ / 149.09 g/mol HNO₃ = 0.5631 mol HNO₃ (limiting) 0.5631 mol HNO₃ × (3 mol Cu(NO₃)₂ / 8 mol HNO₃) × 149.62 g/mol Cu(NO₃)₂ = 93.8 g Cu(NO₃)₂ Therefore, the best choice is B: HNO₃, 93.8 g.

426 G Cu Are Combined with 84