Question

An aqueous solution containing 15.0 g of NaOH is mixed with an aqueous solution containing 15.0 g of H2SO4. Is sodium hydroxide or sulfuric acid the limiting reactant?

   H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

Answer 1

Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol


mass(H2SO4)= 15.0 g

use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(15 g)/(98.09 g/mol)
= 0.1529 mol

Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol


mass(NaOH)= 15.0 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(15 g)/(40 g/mol)
= 0.375 mol
Balanced chemical equation is:
H2SO4 + 2 NaOH ---> Na2SO4 + 2 H2O


1 mol of H2SO4 reacts with 2 mol of NaOH
for 0.1529 mol of H2SO4, 0.3059 mol of NaOH is required
But we have 0.375 mol of NaOH

so, H2SO4 is limiting reagent

Answer: sulphuric acid