Question

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 12.9 g of sodium chloride is produced from the reaction of 19.7 g of hydrochloric acid and 40.8 g of sodium hydroxide, calculate the percent yield of sodium chloride. Round your answer to 3 significant figures.

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)= 19.7 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(19.7 g)/(36.46 g/mol)

= 0.5403 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)= 40.8 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(40.8 g)/(40 g/mol)

= 1.02 mol

Balanced chemical equation is:

HCl + NaOH ---> NaCl + H2O

1 mol of HCl reacts with 1 mol of NaOH

for 0.5403 mol of HCl, 0.5403 mol of NaOH is required

But we have 1.02 mol of NaOH

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

According to balanced equation

mol of NaCl formed = (1/1)* moles of HCl

= (1/1)*0.5403

= 0.5403 mol

use:

mass of NaCl = number of mol * molar mass

= 0.5403*58.44

= 31.58 g

% yield = actual mass*100/theoretical mass

= 12.9*100/31.58

= 40.9 %

Answer: 40.9 %