I’ll get back to u on this
Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of
= 3.00 mole
Moles of
= 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 2 moles of
react with 1 mole of 
So, 3.00 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is, 
(g solute/g solution)*100 = % mass/mass
30 g / 400 * 100
0,075 * 100
= 7,5% w/w
hope this helps!
H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>