Given the reaction: H2SO4 = 2NaOH -> NaSO4 = 2H20.
Determine the # of moles of sulfuric acid id needed to completely react with 2.3 L of 0.5 M of NaOH. Then determine the concentration of the solution if 3.0 L of it are used instead.
avoid typos in your formula:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
2.3L of .5M NaOH contains 1.15 moles NaOH
Since 1 mole H2SO4 reacts with 2 moles NaOH, you will use only 0.575 moles of H2SO4.
The final solution will contain the leftover H2SO4 and a quite soluble Na2SO4. Not sure what the question is getting at.
To determine the number of moles of sulfuric acid (H2SO4) needed to completely react with a given amount of sodium hydroxide (NaOH), you can use the balanced chemical equation:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
In the equation, the stoichiometry tells us that one mole of sulfuric acid reacts with two moles of sodium hydroxide.
First, calculate the number of moles of NaOH present in 2.3 L of a 0.5 M solution:
Molarity (M) = moles of solute / liters of solution
moles of NaOH = Molarity x liters of solution
= 0.5 mol/L x 2.3 L
= 1.15 mol
Since the stoichiometry shows that one mole of H2SO4 reacts with two moles of NaOH, you would need half the number of moles of sulfuric acid to react completely:
moles of H2SO4 = 1.15 mol / 2
= 0.575 mol
So, 0.575 moles of sulfuric acid are needed to completely react with 2.3 L of 0.5 M NaOH.
Next, let's determine the concentration of the solution if 3.0 L of it are used instead.
Using the same approach, calculate the number of moles of NaOH in 3.0 L of the solution:
moles of NaOH = Molarity x liters of solution
= 0.5 mol/L x 3.0 L
= 1.5 mol
Since the stoichiometry remains the same, you would still need half the number of moles of sulfuric acid to react completely:
moles of H2SO4 = 1.5 mol / 2
= 0.75 mol
Therefore, if 3.0 L of the solution are used instead, the concentration remains the same, 0.5 M, and you would need 0.75 moles of sulfuric acid to completely react with it.