What mass of nitrogen is required to react with 16 grams of oxigen?

Answer 1

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

Answer 2

The balanced chemical equation for the reaction between nitrogen and oxygen is:

2N2 + O2 -> 2NO

Using the molar masses, we find that 28 g of nitrogen reacts with 32 g of oxygen. Therefore, 28 grams of nitrogen is required to react with 16 grams of oxygen.

Answer 3

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

Answer 4

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.

Answer 5

N2+3H2 --->2NH3 3mol of H2 give 2mol NH3. So 16.5mol of H2 give 11mol of NH3

Answer 6

The minimum number of moles of 02 that are needed to completely react with 16 moles of NH3 is 12 moles oxygen.

Answer 7

For the reaction:
N2 + O2 = 2 NO
8 g oxygen are necessary.

Answer 8

192 g oxygen are needed.

Answer 9

85,25 g oxygen are needed.

Answer 10

16 g O2 correspond to 0,5 mol.