0
300 moles of nitrogen equals 8404,2 g.
The density of nitrogen at 20 oc and 1 at is 1,251 g/cm3.
8404,2 g nitrogen equals 6 718 L.
1 kL = 1 000 L
So the volume of 300 moles N2 is 6,718 kL.
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoWithout information on temperature and pressure, it is not possible.
Add your answer:
Earn +20 pts
What mass of NH3 is produced when 1.20 mol of N2 react completely in the following equation N2 plus 3H2 2NH3?
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
If 5.0 moles of NH3 are produce how many moles of N2 must have been used?
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
How many moles of N2 are in 50g of it?
50g/28g= 1.7857 moles
How many moles of N2 are produced by the decomposition of 1.20 mol of NaN3?
The number of atoms is 45,166.10e23.
What is the pressure inside a 10.0 L flask containing 14.2 g of N2 at 26 degree celsius. Please show your steps. Iwant to see if I set my problem up correctly. Thank you?
Use this formula. PV = nRT. Need to convert grams N2 to moles N2 and temperature Celsius to temperature Kelvin( add 273.15 )14.2 grams N2 ( 1mole N2/28.02 grams) = 0.507 moles N226 degrees C = 299.15 Kelvin(pressure)(10.0 liter) = (0.507 moles N2)(0.08206 Latm/molK)(299.15 K)= 1.24 atmospheres pressureThank you so much for your answer. I did end up figuring it out on my own.
How many moles of H2are needed to react with 0.90moles N2?
I assume you mean this reaction. N2 + 3H2 --> 2NH3 0.90 moles N2 (3 moles H2/1 mole N2) = 2.7 moles hydrogen gas needed =====================
How many moles of nitrogen are required to convert 9 mol hydrogen gas to ammonia gas?
9H2 + 3N2 ------> 6NH3 3 moles of N2 would be required.
What mass of NH3 is produced when 1.20 mol of N2 react completely in the following equation N2 plus 3H2 2NH3?
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
If 5.0 moles of NH3 are produce how many moles of N2 must have been used?
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
How many moles of N2 reacted if 0.40 mole NH3 is produced?
2 x 0.60 = 1.2 the reaction is N2 + 3H2 -> 2NH3 (1 mole of nitrogen N2 give 2 moles of NH3)
How do you find if you have 14.5 moles of N2 how many moles of H2 are theoretically needed to produce 22.5 moles of NH3 according to the reaction N2 plus 3H2 2NH3?
N2 + 3H2 -> 2NH3 If you have moles produced you can do it this way. 22.5 moles NH3 (3 moles H2/2 moles NH3) = 33.8 moles H2 needed -----------------------------------
How many moles of N2 are in 50g of it?
50g/28g= 1.7857 moles
How many moles of ammonia gas are produced by the reaction of 3 mole of nitrogen gas with excess hydrogen gas N2 H2 ---- NH3?
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
How many moles of N are there in 0.189 grams of N20?
The answer is 0,0043 moles of N2.
How many grams of NH3 can be produced from 12.11 g H2?
N2(g) + 3H2-> 2NH3(g) This is the balanced equation Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important. moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
How many moles of nitrogen and hydrogen are needed to get 10 moles of ammonia?
The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2
How many moles of N2 are produced by the decomposition of 1.20 mol of NaN3?
The number of atoms is 45,166.10e23.
