Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH.
H2SO4(aq) + 2 NaOH(aq) -> 2H2O(l) + Na2SO4(aq).
Suppose a beaker contains 35.0mL of 0.175M H2SO4. How many moles of NaOH are needed to react completely with sulfuric acid?
Thank you
You have 0.0350 L x 0.175M = about 0.006125 mols H2SO4. It will take 2x that for mols NaOH
To find out how many moles of NaOH are needed to react completely with sulfuric acid, we can use the balanced chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction between H2SO4 and NaOH is:
H2SO4(aq) + 2NaOH(aq) -> 2H2O(l) + Na2SO4(aq)
From this equation, we can see that 1 mole of sulfuric acid (H2SO4) reacts with 2 moles of sodium hydroxide (NaOH).
To calculate the number of moles of H2SO4 in the beaker, we can use the formula:
moles = concentration (M) × volume (L)
Given that the volume of the acid is 35.0 mL, we need to convert it to liters by dividing by 1000:
Volume (L) = 35.0 mL ÷ 1000 mL/L = 0.035 L
Now, we can calculate the number of moles of H2SO4 using the concentration:
moles of H2SO4 = concentration (M) × volume (L)
= 0.175 M × 0.035 L
= 0.006125 moles of H2SO4
Since the stoichiometric ratio between H2SO4 and NaOH is 1:2, we can determine that twice the number of moles of NaOH is needed. Therefore:
moles of NaOH = 2 × moles of H2SO4
= 2 × 0.006125
= 0.01225 moles of NaOH
So, 0.01225 moles of NaOH are needed to react completely with the given amount of sulfuric acid in the beaker.