Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O . If 2.88g of water is produced from the reaction of 15.7g of hydrochloric acid and 9.12g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

PLEASE make sure there are the right number of significant digits

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)= 9.12 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(9.12 g)/(39.998 g/mol)

= 0.228 mol

Molar mass of HCl,

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

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 15.7 g

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(15.7 g)/(36.458 g/mol)

= 0.4306 mol

Balanced chemical equation is:

NaOH + HCl ---> H2O + 2 NaCl

1 mol of NaOH reacts with 1 mol of HCl

for 0.228011 mol of NaOH, 0.228011 mol of HCl is required

But we have 0.430633 mol of HCl

so, NaOH is limiting reagent

we will use NaOH 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 NaOH

= (1/1)*0.228011

= 0.228011 mol

mass of H2O = number of mol * molar mass

= 0.228*18.02

= 4.108 g

This is theoretical mass of H2O produced

% yield = actual mass*100/theoretical mass

= 2.88*100/4.108

= 70.1 %

Answer: 70.1 %