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

Learn more about excess reagent here:

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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.

the answer is the first option

explanation:

just googled it m8. not that hard

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explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.