You want to neutralize 6 mol of sulfuric acid with 6.0 M sodium hydroxide. How much 6.0 M sodium hydroxide must you add to react exactly with the sulfuric acid?
2NaOH(aq)+H2SO4(aq)-->2H2O(l)+Na2SO4(aq)
2 moles sodium hydroxide for every 1 mol sulfuric acid.
I do not know what steps to take to solve this problem from here.
PLEASE HELP
You did the most important first two. The next one is the easy one.
So you want to neutralize 6 mols H2SO4, that will take 12 mols NaOH as you correctly stated.
Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.
To solve this problem, you need to determine the amount of 6.0 M sodium hydroxide (NaOH) required to react completely with 6 mol of sulfuric acid (H2SO4).
First, let's calculate the number of moles of sodium hydroxide required. The balanced chemical equation shows that 2 moles of NaOH are needed for every 1 mole of H2SO4. Since you have 6 moles of H2SO4, you will need:
6 mol H2SO4 × (2 mol NaOH / 1 mol H2SO4) = 12 mol NaOH
This means that 12 moles of NaOH would be required to react completely with 6 moles of sulfuric acid.
Now, we need to convert the moles of NaOH to the volume of 6.0 M NaOH solution. The concentration of 6.0 M NaOH can be interpreted as 6 moles of NaOH dissolved in 1 liter of solution.
To calculate the volume, you can use the equation:
Volume (in liters) = Moles of solute (NaOH) / Concentration (6.0 M)
Volume = 12 mol NaOH / 6.0 M = 2 liters NaOH
Therefore, you would need to add 2 liters of 6.0 M sodium hydroxide solution to react exactly with 6 mol of sulfuric acid.