Question

Aqueous hydrochloric acid

HCl

reacts with solid sodium hydroxide

NaOH

to produce aqueous sodium chloride

NaCl

and liquid water

H2O

. If

0.314g

of water is produced from the reaction of

0.73g

of hydrochloric acid and

1.3g

of sodium hydroxide, calculate the percent yield of water.

Be sure your answer has the correct number of significant digits in it.

Answer 1

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 0.73 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.73 g)/(36.458 g/mol)
= 2.002*10^-2 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)= 1.3 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(1.3 g)/(39.998 g/mol)
= 3.25*10^-2 mol
Balanced chemical equation is:
HCl + NaOH ---> H2O + NaCl


1 mol of HCl reacts with 1 mol of NaOH
for 2.002*10^-2 mol of HCl, 2.002*10^-2 mol of NaOH is required
But we have 3.25*10^-2 mol of NaOH

so, HCl is limiting reagent
we will use HCl 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 = (1/1)* moles of HCl
= (1/1)*2.002*10^-2
= 2.002*10^-2 mol


use:
mass of H2O = number of mol * molar mass
= 2.002*10^-2*18.02
= 0.3607 g

% yield = actual mass*100/theoretical mass
= 0.314*100/0.3607
= 87.0 %
Answer: 87 %