A 0.885 M solution of sulfuric acid (H2SO4) is to be reacted with a 0.552 M solution of sodium hydroxide to make sodium sulfate and water. How many L of the H2SO4 solution will react with 0.842 L of the NaOH solution? Give your answer to 3 decimal places.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols H2SO4 = grams/molar mass
Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH.
Then M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L.
To determine the number of liters of the H2SO4 solution that will react with the NaOH solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
From the balanced equation, we can see that one mole of H2SO4 reacts with two moles of NaOH.
Step 1: Convert the given volumes to liters.
0.842 L of NaOH solution
Step 2: Convert the volumes to moles using the molarity (M) and volume (L) relationship.
moles of NaOH = Molarity of NaOH x Volume of NaOH
moles of NaOH = 0.552 M x 0.842 L
Step 3: Use the stoichiometry ratio to relate the moles of NaOH to the moles of H2SO4.
Since the stoichiometry ratio between NaOH and H2SO4 is 2:1, we will divide the moles of NaOH by 2 to get the moles of H2SO4.
moles of H2SO4 = moles of NaOH / 2
Step 4: Convert the moles of H2SO4 to volume using the molarity (M) and volume (L) relationship.
Volume of H2SO4 = moles of H2SO4 / Molarity of H2SO4
Step 5: Calculate the volume of H2SO4 by substituting the values obtained from previous steps.
Volume of H2SO4 = (moles of NaOH / 2) / 0.885 M
Step 6: Lastly, round the answer to 3 decimal places.
Now you can plug in the values into the equation to find the answer:
Volume of H2SO4 = (0.552 M x 0.842 L / 2) / 0.885 M
= 0.260 L (rounded to 3 decimal places)
Therefore, 0.260 L of the H2SO4 solution will react with 0.842 L of the NaOH solution.