Answer: (D) 0.025 mol [tex]O_2[/tex]
Explanation:
The given balanced equation is :
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
According to stoichiometry :
4 moles of ammonia [tex](NH_3)[/tex] reacts with = 5 moles of oxygen [tex](O_2)[/tex]
Thus 0.30 moles of ammonia [tex](NH_3)[/tex] reacts with = [tex]\frac{5}{4}\times 0.30=0.375[/tex] moles of oxygen [tex](O_2)[/tex]
Now as given moles of oxygen are more than the required amount, oxygen is the excess reagent.
moles of oxygen [tex](O_2)[/tex] left = 0.40 - 0.375 = 0.025
Thus 0.025 mol [tex]O_2[/tex] excess reagent remains.
The amount of excess reagent remaining when 0.30 mole NH3 reacts with 0.40 mole O2 is 0.025 mole of O2
An excess reagent is a reactant which is left and can keep reacting after whole reaction.
A limiting reagent is a reactant that is totally consumed in the whole process of the reaction.
The chemical equation for the reaction is represented as:
[tex]\mathbf{4NH_3 + 5O_2 \to 4NO + 6H_2O}[/tex]
From the above reaction,
4 moles of ammonia reacts with 5 moles of oxygen molecule;
If 0.30 moles of ammonia reacts, then the numbers of moles of oxygen required can be computed as:
[tex]\mathbf{= \dfrac{0.3 \times 5}{4}}[/tex]
= 0.375 moles of oxygen.
But, we are given that 0.40 moles of oxygen is present.
Therefore, the amount of oxygen in excess is
= 0.40 moles - 0.375 moles
= 0.025 moles of O2
Learn more about excess reagent here:
https://brainly.com/question/25680776
