Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
Second orbital can hold 8 electrons
Answer:
Rate expression has been given below
Explanation:
According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.
So, rate of disappearance of both A and B are one half of rate of appearance of B
Hence rate expression can be represented as:
![Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%5Cfrac%7B-%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B-%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where ![\frac{-\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
is rate of disappearance of A, ![\frac{-\Delta [B]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D)
is rate of disappearance of B and ![\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
rate of appearance of C
B.
Delta G is negative and Delta S is positive.
