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!
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~ :)