4 NH3 + 5 O2 ---> 4 NO + 6 H2O
moles NH3 used = 36.3 g x 1 mole/17 g = 2.14 moles
moles O2 needed = 2.14 moles (note a 1mole to 1mole ratio of O2 to NH3 in balanced equation)
grams O2 needed = 2.14 moles x 32g/mol = 68.48 grams needed
20
this is because you know that it takes 5 moles of o2 to react with 4 moles of nh3
now 02 equals 32 multiply that by 5 and that equals 160
nh3 equals 18 multiplied by 4 is 72
divide them (160/72)
multiply your answer by 288 which you got by multiplying nh3 by 16
after doing all of this you should get 639.9999 you simply divide that by 32
and you get 20
2 N2 (g) + O2 (g) ⇌ 2 N2O (g)16 g O2 x 1 mole O2/32 g = 0.5 moles O2 present
0.5 moles O2 x 2 moles N2/mole O2 = 1.0 mole N2 required
Mass of N2 required = 1.0 mole x 28 g/mole = 28 grams N2 required.
N2+3H2 --->2NH3 3mol of H2 give 2mol NH3. So 16.5mol of H2 give 11mol of NH3
The minimum number of moles of 02 that are needed to completely react with 16 moles of NH3 is 12 moles oxygen.
For the reaction:
N2 + O2 = 2 NO
8 g oxygen are necessary.
192 g oxygen are needed.
85,25 g oxygen are needed.
16 g O2 correspond to 0,5 mol.
5.6 g
8.62
69
why Na does not react with nitrogen
Thevolume is 24,12 mL.
To form ammonia, reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 1.13 moles nitrogen is required.
How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide?
oxigen
11.5 g NO2
g
16g
8.62
160...cant quite grasp HOW though
69
why Na does not react with nitrogen
Nitrogen is the limiting reactant and 4.15g of ammonia are produced.
The formula N2O5 shows that there are 2/5 as many nitrogen atoms as oxygen atoms in the compound. Therefore, the number of nitrogen atoms required is (2/5)(7.05 X 1022) or 2.82 X 1022 atoms. The gram atomic mass of nitrogen is 14.0067 and, by definition, consists of Avogadro's Number of atoms. Therefore, the mass of nitrogen required to react with the specified amount of oxygen to produce the specified compound is 14.0067 [(2.82 X 1022)/(6.022 X 1023] or 0.656 grams of nitrogen, to the justified number of significant digits.
Thevolume is 24,12 mL.