For the reaction NH3 + O2 → NO + H2O?

Question

How many moles of NO are produced from the complete reaction of 4.78 moles of O2?

Answer 1

First you must balance the equation:
4NH3 + 5O2 -> 4NO + 6H2O
there are 4N, 12H and 10O on each side
you have 4.78 moles of O2 you need to divide it by the number of moles of O2 that react and multiply it by the number of moles of NO produced.
4.78 / 5 x 4 = 3.824moles
Hope this helps

Answer 2

First you have to balance the reaction to determine the mole ratios of the reactants and products.

The balanced reaction is:

4NH3 + 5O2 --> 4NO + 6H2O

So the molar ration is 5 to 4

So the conversion factor would be 4/5. Remember the moles of the unknown is in the numerator.

And the answer would be 4.78 x 4/5 or 3.824 moles of NO. And you should round off to 3.82 moles to keep the same number of significant digits.

Answer 3

you have have been given to have a balanced equation first. 4NH3 + 5O2 --> 4NO + 6H2O grams A --> moles A --> moles B --> grams B A) molNH3 = 80 5.15g/17.0g/mol ~ 5.0 mol NH3 mol O2 = 5.0 mol NH3 x 5 mol O2/4mol NH3 (from the coefficients interior the balanced equation) = 6.25 mol O2 g O2 = 6.25mol x 32.0 g/mol = 200 g O2 B) persist with an identical technique as in A.

Answer 4

4NH3 + 5O2 ==> 4NO + 6H2O
4.78 moles of O2 its a 5:4 ratio
so
4.78 mol / 5 = x mol/ 4
therefore there are 3.824 mol of NO produced

Answer 5

4NH3 + 5O2 --> 4NO + 6H2O

4.78moles O2 x (4moles NO / 5moles O2) = 6.22moles NO

Answer 6

3.82 moles NO