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. which means there is 12 mol sodium hydroxide to 6 mol sulfuric acid.
I do not know what steps to take to solve this problem from here.
-How do I convert from moles of aulfuric acid to moles of sodium hydroxide?
I answered this question earlier.
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To convert from moles of sulfuric acid to moles of sodium hydroxide, you can use the stoichiometric ratio from the balanced equation.
From the balanced equation, we know that 2 moles of sodium hydroxide (NaOH) react with 1 mole of sulfuric acid (H2SO4). This means that the stoichiometric ratio between NaOH and H2SO4 is 2:1.
In your case, you have 6 moles of sulfuric acid (H2SO4) that you want to neutralize. To determine the amount of sodium hydroxide needed, you can set up a simple ratio and solve for the unknown quantity:
6 moles H2SO4 : x moles NaOH = 1 mole H2SO4 : 2 moles NaOH
By using the concept of a proportion, you can cross-multiply and solve for x:
6 moles H2SO4 * 2 moles NaOH = x moles NaOH * 1 mole H2SO4
12 moles NaOH = x moles NaOH
Therefore, you need 12 moles of sodium hydroxide (NaOH) to completely neutralize 6 moles of sulfuric acid (H2SO4).