moles help?

Question

i just cant seem to get to grips with moles can someone walk me through this question

In a titration, 25cm3 of 0.01M potassium hydroxide is pipetted into a conical flask. Sulphuric acid is run in from the burette and it is found that 19.8 cm3 were required to react exactly. Calculate the concentration of a solution of sulphuric acid

H2SO4(aq) + 2HNO3(aq) to K2O(l) + CO2(g)

Answer 1

1st, the equation should have KOH (potassuim hydroxide) in it. To be reacted exactly, then it should be an acid-base neutralization reaction:
H2SO4(aq) + KOH(aq) ---> H2O(l) + K2SO4(aq)
Balance it:
H2SO4(aq) + 2KOH(aq) ---> 2H2O(l) + K2SO4(aq)


Contants to know:
1 cm3 = 1 mL
1000 mL = 1 L
1 M = 1 mol/L

Conversions:
25 cm3 KOH = 25 mL KOH = 0.025 L KOH
If there are 0.025 L of a .01M KOH solution, you can find moles of the KOH:
0.025 L * .01 mol/L = 0.00025 mol KOH

19.8 cm3 H2SO4 reacted = 19.8 mL H2SO4 = 0.0198 L H2SO4

So you have to find the mol of H2SO4 to find concentration (M = mol/L)

If you have .00025 mol KOH, you can use the mole fraction in the balanced equation:
H2SO4(aq) + 2KOH(aq) ---> 2H2O(l) + K2SO4(aq)
2 moles of KOH are needed to react with every 1 mole of H2SO4 (mole fraction: 1 mol H2SO4/2 mol KOH)

If you have .00025 mol KOH, multiply it by the mole fraction:
0.00025 mol KOH * 1 mol H2SO4/2 mol KOH = 0.000125 mol H2SO4

Now you have the moles and liters of H2SO4.
Calculate concentration (molarity = M = mol/L)
M H2SO4 = 0.000125 mol H2SO4 / 0.0198 L H2SO4 = 0.006313 M H2SO4

Hope I helped ^^ please rank :)

Answer 2

First off... the reaction you have listed is not potassium hydroxide reacting with sulfuric acid... The reaction from the question should say be (pretty much): H2SO4 + 2KOH -> 2H2O + K2SO4.

That being said the first step to solving this is to realize that the amount of moles of potassium hydroxide(KOH) is twice that sulfuric acid(H2SO4), as is shown in the equation.

The second step is to calculate the amount of moles of KOH used. The volume of KOH used is 25cm3 = 25ml = 0.025L. The concentration of KOH = 0.01M = 0.01 moles/L. If you multiply the volume by the concentration you get the amount of moles of KOH used: 0.01 moles/L * 0.025L = 0.00025moles.

Since we know that 0.00025 moles of KOH was used we know that half that amount of H2SO4 was used. So to get the concentration of H2SO4 divide the number of moles of H2SO4 by the volume of H2SO4 (0.0198L).

Answer 3

Skip to the below part...this wont work in this situation.
It is my understanding that this is a basic titration problem. You can use Molar(a)*Volume(a) = Molar(b)*Volume(b)

25cm^3 (.01) = x (19.8cm^3)

x=.012626 which is M of sulfuric acid

The number of moles shouldn't really matter because you are finding the overall M in the solution of just sulfuric acid.
______________________________________________
Forget the above here is my explanation:
First off we write the stoichiometry
H2S04 + 2KOH --> K2SO4 + 2H20
We find moles of KOH = (.01moles/L)(.025L) = .00025 moles
We know that 2 moles of KOH will react exactly with 1 mole of H2S04, so moles of H2SO4 is .00025/2 = 1.25E-4

1.25E-4/.0198L = .006313 moles per liter or .006313M

Answer 4

H2SO4(aq) + 2KOH(aq) ===> K2SO4(aq) + 2H2O

Let the sulfuric acid solution be called SA. Let the potassium hydroxide solution be called PH. Let cm3 be called mL.

25mLPH/19.8mLSA x 0.01molKOH/1000mLPH x 1molH2SO4/2molKOH x 1000mLSA/1LSA = 0.0063 mol/L

Answer 5

OK......WHAT THE FFFFFFF!!

Answer 6

WTF??!!! I don't get anything on here LoL sorry can't help you