The first thing to do is to make sure we have a balanced chemical equation, which was given in the problem. This is an acid-base reaction but it's still just a stoichiometry problem in that we have to worry about the moles of each substances.
Since we are given the mass of NaOH, we can find the moles of NaOH. ONce we know the moles, we can use the mole ratio from the balanced chemical equation to see that for every 1 mole of H2SO4, we will need 2 moles of NaOH.
We can set up our calculation, paying careful attention to theunitsto make sure they cancel out and that we end up with a volume. Remember that molarity (M) is moles of solute per liter of solution. Rewriting as mol/L will help us see how we need to use it in the equation.
2.5 kg NaOH(1000 g/1 kg)(1 mol NaOH/40 g)(1 mol H2SO4/2 mol NaOH)(1 L/3.00 mol H2SO4)
Once we cancel out our units, we get L which is a volume and what the problem is asking for.
The answer is 10.4 L H2SO4
The balanced reaction is
H2SO4+2NaOH --> Na2SO4+2H2O
Mass of NaOH is given as 2.5 kg = 2500 g
And molar mass of NaOH is 40 g/mol
Moles = mass/molar mass
moles of NaOH = 2500 g/ (40g/mol) = 62.5 mol NaOH
If we look at the reaction, 1 mol of H2So4 requries 2 mol NaOH for neutrilization. That is the molar artio is 1 mol H2SO4 : 2 mol NaOH
So moles of H2So4 requries for neutrilization of 62.5 mol NaOH would be
62.5 mol NaOh * (1 mol H2SO4/ 2 mol NaOH)
31.25 mol H2SO4.
Here the concentration of H2sO4 IS 3 M
Molarity = moles/volume(liter)
Voilume(liter) = Moles/molarity
Volume of H2SO4 = 31.25 mol H2SO4/ 3 M = 10.42 L