Question

33. When 45.0 g NaOH and 40.0 g H2SO4 are mixed and reacted, which is the limiting reagent? H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

Answer 1

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)= 45.0 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(45 g)/(40 g/mol)

= 1.125 mol

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)= 40.0 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(40 g)/(98.09 g/mol)

= 0.4078 mol

Balanced chemical equation is:

2 NaOH + H2SO4 ---> Na2SO4 + 2 H2O

2 mol of NaOH reacts with 1 mol of H2SO4

for 1.125 mol of NaOH, 0.5625 mol of H2SO4 is required

But we have 0.4078 mol of H2SO4

so, H2SO4 is limiting reagent

Answer: H2SO4