Posted by **Sarah** on Friday, April 15, 2011 at 11:42am.

You have 315.00g of sodium hydroxide. How many liters of dangerous 12.0M sulfuric acid (H2SO4) can you neutralize with your supply?

__NaOH(s) +__ H2SO4(aq) -> __Na2SO4 + __H2O(l)

I know the answer is .43, but I don't know how to get there. Thanks!

- More Solutions Chem -
**Jai**, Friday, April 15, 2011 at 12:08pm
first we balance the given chemical equation, thus:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

note that we need to balance a chemical equation because we need their stoichiometric coefficients (number before the chemical formula of a compound or element) to make ratios.

since we are given the mass of NaOH, we first find its molar mass. from the periodic table of elements, masses of Na, O and H are

Na = 23 , O = 16 , H = 1 thus

23 + 16 + 1 = 40 g/mol

therefore, we start with the given. and we solve for the number of moles of H2SO4 produced for a given amount of NaOH:

315 g NaOH * (1 mol NaOH / 40 g NaOH) * (1 mol H2SO4 / 2 mol NaOH) = 3.9375 mol H2SO4

since we need to find the volume (in liters) of H2SO4 given the molarity, recall that molarity (concentration) is given by

M = n / V

where n = moles and V = volume (in L)

substituting,

12 = 3.9375/V

V = 3.9375/12

V = 0.328 L

*are you sure with the answer 0.43 ?

hope this helps~ :)

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