Consider the formation of dinitrogen pentoxide from nitrogen dioxide and oxygen
ID: 1008289 • Letter: C
Question
Consider the formation of dinitrogen pentoxide from nitrogen dioxide and oxygen 4NO_2 (g)+ O_2(g) rightarrow 2 N_2O_5 (g) a. If 40.0L of nitrogen dioxide reacts with 40.0L of oxygen, what is the volume of dinitrogen pentoxide is produced? Assume that all reactants and products are under the same conditions. b. What is the total volume of all gases after the reaction in )? (at STP). c. Like parts a and b, starting with 50.0L of nitrogen dioxide, but only 10.0L of oxygen calculate the total volume at the end (STP). d. Like, starting with 60.0L of nitrogen dioxide and 16.0L of oxygen, calculate the total volume at the end (STP).Explanation / Answer
(a): Volume of 1 mol of a gas at STP is 22.4 L
The balanced chemical reaction is
4 NO2(g) + O2(g) ------- > 2N2O5(g)
4 mol, ----- 1 mol, ------------ 2 mol
4x22.4L, 22.4 L, ------------ 2x22.4 L
40.0 L NO2 that will react with the moles of O2 = 40.0 L NO2 x (22.4 L O2 / 4x22.4 L NO2) = 10.0 L O2
i.e 40.0 L NO2 reacts with only 10.0 L O2.
Hence NO2 is completely exhausted and is the limiting reactant. Hence amount of NO2 will decide the amount of product formed.
Hence volume N2O5 formed = 40.0 L NO2 x (2x22.4 L N2O5 / 4x22.4 L NO2) = 20.0 L N2O5 (answer)
(b):
NO2 is completely exhausted.
volume of O2 remain unreacted = 40.0 L - 10.0 L = 30.0 L
Volume of N2O5 formed = 20.0 L
Hence total volume of all gases = 0 L + 30.0 L + 20.0 L = 50.0 L (answer)
(c):
Since 40.0 L NO2 reacts with 10.0 L O2, now O2 is completely exhausted and becomes limiting reactant. Also the volume of O2 will decide the volume of N2O5 formed
Hence volume of NO2 remained unreacted = 50.0 L - 40.0 L = 10.0 L
Volume of N2O5 formed = 10.0 L O2 x (2x22.4 L N2O5 / 1x22.4 L O2) = 20.0 L N2O5
Hence total volume at the end
= 10.0 L + 0 L + 20.0 L = 30.0 L (answer)
(d):
60.0 L NO2 that will react with the volume of O2 = 60.0 L NO2 x (1x22.4 L O2 / 4x22.4 L NO2)
= 15.0 L O2
Hence 60.0 L NO2 is completely exhausted and will decide the amount of N2O5 produced.
Volume of excess O2 remain unreacted = 16.0 L - 15.0 L = 1.0 L
Volume of N2O5 formed = 60.0 L NO2 x (2x22.4 L N2O5 / 4x22.4 L O2) = 30.0 L N2O5.
Hence total volume at the end
= 0 L + 1.0 L + 30.0 L
= 31.0 L (answer)