Suppose 100 mL of NO at STP is mixed with 400 mL of O2 at STP.
2 NO (g) + O2 (g)---> 2 NO2 (g)
After the reaction goes to completion, what is the partial pressure of NO2 in the resulting mixture of gases at STP?
I know the relation between pressure and volume is P1V1=P2V2, but I am not sure how to use it in this question.
3 answers
I used PV=nRT for both the gases but I am not getting the right answer. Please help
I believe the way to approach this problem is through mole fraction. With all materials in the gaseous phase one can use volume as mols; first determine the limiting reagent. 100 mL NO will give you 100 x (2 mol NO2/2 mols NO) = 100 mL NO if we had all of the O2 needed.
For 400 mL O2 x (2 mols NO2/1 mol O2) = 800 mL NO2.
This means NO is the limiting reagent. Therefore, we will use 100 mL NO2 and there will be zero left, we will use 50 mL O2 (100 mL NO x (1 mol O2/2 mol NO) which will leave 350 mL O2 after reaction (400 mL-50 mL = 350 mL) and we will have 100 mL NO formed.
mols NO = 0
mol O2 = 0.350/22.4 = ?
mol NO2 = 0.1/22.4 = ?
ntotal = mols O2 + mols NO2.
mol fraction NO2 = XNO2 = nNO2/total mols
Then pNO2 = XNO2*Ptotal.
Ptotal is 760 mm. Solve for pNO2.
Note that the same answer is obtained by using volume as mols; i.e.,
100 mL + 350 mL = 450 mL.
XNO2 = 100/450 = 0.222 etc.
For 400 mL O2 x (2 mols NO2/1 mol O2) = 800 mL NO2.
This means NO is the limiting reagent. Therefore, we will use 100 mL NO2 and there will be zero left, we will use 50 mL O2 (100 mL NO x (1 mol O2/2 mol NO) which will leave 350 mL O2 after reaction (400 mL-50 mL = 350 mL) and we will have 100 mL NO formed.
mols NO = 0
mol O2 = 0.350/22.4 = ?
mol NO2 = 0.1/22.4 = ?
ntotal = mols O2 + mols NO2.
mol fraction NO2 = XNO2 = nNO2/total mols
Then pNO2 = XNO2*Ptotal.
Ptotal is 760 mm. Solve for pNO2.
Note that the same answer is obtained by using volume as mols; i.e.,
100 mL + 350 mL = 450 mL.
XNO2 = 100/450 = 0.222 etc.
Is the total pressure 1 atm because the questions says STP?
0 answers