Textbook Problem M11 (C)

[KMcFarland_2L]

To find the amount of reactant left, you have to find the amount that was created in the first reaction, but in the problem, oxygen is the limiting reactant, however to find the product created in the first reaction the amount of phosphorus in grams is used to find the amount of product. Why would you use the non-limiting reactant for this situation?

[Nathan Tong 3G]

I believe that in the first reaction with P₄, oxygen is not the limiting reactant, as there are fewer moles of P₄ than O₂. In fact, there is an excess of oxygen that can react with P₄O₆ to create P₄O₁₀. However, there is not enough O₂ left over from the first reaction to react with all the resulting P₄O₆ in the second reaction. So to sum it up, you would need to find out how much O₂ is left over from the first reaction, and use the extra O₂ to find out how much P₄O₁₀ you can make when it reacts with P₄O₆. Hope that makes sense.

[Thomas Vu 1A]

Yea so oxygen is the limiting reactant for the second equation. Just to check though since the answer key didn't say anything for this problem, did anyone else get 5.79g P4O10 for b) and 5.74 g P4O6 for c). I did this problem pretty quick so I'm not sure if those are right. thanks