What amount of the excess reagent remains when 0.30 mol NH3 reacts with 0.40 mol O2 to produce NO and H2O? 4NH3 +502 + 4NO + 6H20 (A) 0.10 mol NH,
(B) 0.10 mol
(C) 0.025 mol NH
(D) 0.025 mol O2

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:

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:

= 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

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(D) 0.025 mol

Explanation:

The given balanced equation is :

According to stoichiometry :

4 moles of ammonia reacts with = 5 moles of oxygen

Thus 0.30 moles of  ammonia    reacts with = moles of oxygen

Now as given moles of oxygen are more than the required amount, oxygen is the excess reagent.

moles of oxygen left = 0.40 - 0.375 = 0.025

Thus 0.025 mol excess reagent remains.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.