Answer:
m of NH3 = 6.46 g
Explanation:
First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:
NH₃ + O₂ ---------> NO + H₂O
Now, let's balance the equation:
4NH₃ + 5O₂ ---------> 4NO + 6H₂O
Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:
MM NH3 = 17 g/mol
MM O2 = 32 g/mol
moles NH3 = 18.1 / 17 = 1.06 moles
moles O2 = 27.2 / 32 = 0.85 moles
Now, let's compare these moles with the theorical moles that the balanced equation gave:
4 moles NH3 --------> 5 moles O2
1.06 moles ----------> X
X = 1.06 * 5 / 4 = 1.325 moles of O2
These means in order to NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.
Now, let's see how many grams in excess we have left after the reaction is complete.
4 moles NH3 --------> 5 moles O2
X moles NH3 ----------> 0.85 moles
X = 0.85 * 4 / 5 = 0.68 moles of NH3
This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:
moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles
Finally the mass:
m = 0.38 * 17
<em>m = 6.46 g of NH3</em>
