Question

ok this problem has made me crazy , 15 % is w\v ok , so i just the m1 , the mass of naoh i don't need the volume , so ill reformulate the problem again , I need the mass of naoh with conc.15% w/v in 176 g h2so4 ok

Answer 1

It still makes no sense to me. Here is the problem as I see it. I don't understand what the 15% has to do with the problem.
What is the mass of NaOH (notice I use caps since naoh means absolutely nothing) required to neutralize 176 g H2SO4.
2NaOH + H2SO4 ==> 2H2O + Na2SO4

moles H2SO4 = 176g/molar mass H2SO4 = ??
Convert to mols NaOH using the coefficients in the balanced equation.
moles NaOH = moles H2SO4 x (2 moles NaOH/1 mole H2SO4) = ??
Then g NaOH = moles NaOH x molar mass NaOH.

mole H2SO4 = 176/98 = 1.796
mole NaOH = 2*1.796 = 3.592
g NaOH = 3.592*40 = 143.7 g.

Answer 2

the problem requires the quantity of NaOH, this is what i don't understand what does it mean by quantity , the volume , the mass ,or the molar concentration ?? it must be the molar concen. of NaOH since i have to use all the things that are give to the exerc. including the 15 % concentration of NaOH