Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
<u>Answer:</u> NO is the limiting reagent in the given reaction and 0.857 moles of 
will be produced.
<u>Explanation:</u>
Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
For the given chemical reaction:
We are given:
Moles of NO = 0.857 mol
Moles of oxygen = 0.498 mol
By stoichiometry of the reaction:
If 2 moles of NO reacts with 1 mole of oxygen gas.
So, 0.857 moles of NO will react with = 
of 
As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, NO is considered as the limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
If 2 moles of NO produces 2 moles of nitrogen dioxide gas.
So, 0.857 moles of NO will produce = 
of
Hence, NO is the limiting reagent in the given reaction and 0.857 moles of will be produced.
It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.
Volume of pure acid = 520*0.13 = 67.6 mL
