how many mole of nog can be produced in the reaction of 3.36 nik nh3(g)and 5.20 mole 02g?
in the 4nh3(g)+ 5o2(g)------->4nog+6h2o(1)
To find out how many moles of NOg can be produced in the reaction, you need to use the stoichiometry of the balanced equation. The balanced equation you provided is:
4NH3(g) + 5O2(g) → 4NOg + 6H2O(l)
From the equation, you can see that the stoichiometric ratio between NH3 and NOg is 4:4, which means that when 4 moles of NH3 react, 4 moles of NOg are produced.
Similarly, the stoichiometric ratio between O2 and NOg is 5:4, meaning that 5 moles of O2 react to produce 4 moles of NOg.
Now, let's use these ratios to solve the problem:
Given:
- Number of moles of NH3 (nik) = 3.36 mol
- Number of moles of O2 = 5.20 mol
We will first determine the limiting reactant, which is the reactant that is completely consumed and limits the amount of product that can be formed. To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio.
NH3:O2 ratio = 4:5
Compare the ratios:
NH3 ratio = 3.36 mol / 4 = 0.84
O2 ratio = 5.20 mol / 5 = 1.04
Since the NH3 ratio is smaller, NH3 is the limiting reactant.
Now, we need to determine how many moles of NOg can be produced using the limiting reactant.
Using the stoichiometric ratio between NH3 and NOg (4:4), we can set up a proportion:
4 mol NH3 / 4 mol NOg = 0.84 mol NH3 / x mol NOg
Solving for x:
x = (0.84 mol NH3 * 4 mol NOg) / 4 mol NH3
x = 0.84 mol NOg
Therefore, when 3.36 mol of NH3 react, 0.84 mol of NOg can be produced.
Note: We do not need to consider the stoichiometric ratio between O2 and NOg because NH3 is the limiting reactant, and once it is completely consumed, the reaction stops, regardless of the excess amount of O2.