Question

How many lbs/day of pure sodium hydroxide are required to neutralize an industrial waste with an acidity equivalent to 20 lbs/day of sulfuric acid? H2SO4 + NaOH = H2O + Na2SO4 (unbalanced equation)

Answer 1

Ans :

The balanced reaction is given as :

H2SO4 + 2NaOH = 2H2O + Na2SO4

So each mole of acid requires to mol of sodium hydroxide.

Mass of acid in one day = 20 lbs = 20 x 453.592 = 9071.85 grams

Number of mol of acid = given mass / molar mass

= 9071.85 / 98.079 = 92.49 mol

So the number of mol of NaOH required = 2 x 92.49

= 184.99 mol

Mass of pure sodium hydroxide required in 1 day = 184.99 x molar mass

= 184.99 x 39.997

= 7399.07 grams

Mass in lbs = 7399.07 / 453.592

= 16.3 lbs

So 16.3 lbs/day of sodium hydroxide will be required.