Question

Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO4) and liquid water (H2O). If 3.37 g of water is produced from the reaction of 13.7 g of sulfuric acid and 20.6 g of sodium hydroxide, calculate the percent yield of water.

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)= 13.7 g

use:

number of mol of H2SO4,

n = mass of H2SO4/molar mass of H2SO4

=(13.7 g)/(98.09 g/mol)

= 0.1397 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)= 20.6 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(20.6 g)/(40 g/mol)

= 0.515 mol

Balanced chemical equation is:

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

1 mol of H2SO4 reacts with 2 mol of NaOH

for 0.1397 mol of H2SO4, 0.2793 mol of NaOH is required

But we have 0.515 mol of NaOH

so, H2SO4 is limiting reagent

we will use H2SO4 in further calculation

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (2/1)* moles of H2SO4

= (2/1)*0.1397

= 0.2793 mol

use:

mass of H2O = number of mol * molar mass

= 0.2793*18.02

= 5.033 g

% yield = actual mass*100/theoretical mass

= 3.37*100/5.033

= 67.0 %

Answer: 67.0 %