What volume of oxygen gas at 320 K and 680 torr will react?

Question

What volume of oxygen gas at 320 K and 680 torr will react completeley with 2.50 L of NO gas at the same temperature and pressure?
2NO(g) + O2(g) 2NO2(g)

Answer 1

Use Ideal Gas Law to get both into moles to solve this.
PV=nRT
0.895 atm x 2.50 L = n x 0.0821 x 320
n = 0.0851 mol NO
From the balanced equation you can see that this would require 0.0426 mol of O2 to completely react with it. Now use Ideal Gas Law again:
0.895 atm x V = 0.0426 x 0.0821 x 320
V = 1.25 L of O2

Answer 2

At the same T and P, equal volumes of gases will have equal numbers of molecules. So the liters can stand in for the moles.

2.50LNO x 1LO2/2LNO = 1.25L O2

Answer 3

you are going to be able to desire to hit upon the moles of O2, mandatory for the reaction VO2 = nO2*R*T/P (nO2 = no of moles O2) Use stochiometry to get moles of O2: 5O2 + C3H8 ----> 3CO2 + 4H20 The no moles of O2 is 5 cases the no of moles of C3H8. locate the no of moles of C3H8 from the gasoline regulation nPr = P*VPr/R*T nO2 = 5*nPr = 5*P*VPr/R*T V = 5*P*VPr/R*T * R*T/P = 5*VPr = 15.0 L In different words, if the temperature and rigidity are the comparable, the quantity ratio is the comparable because of the fact the mole ratio,