Question

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 24. g of hydrochloric acid is mixed with 34.8 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

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)= 24.0 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(24 g)/(36.46 g/mol)

= 0.6583 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)= 34.8 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(34.8 g)/(40 g/mol)

= 0.87 mol

Balanced chemical equation is:

HCl + NaOH ---> H2O + NaCl

1 mol of HCl reacts with 1 mol of NaOH

for 0.6583 mol of HCl, 0.6583 mol of NaOH is required

But we have 0.87 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)*0.6583

= 0.6583 mol

use:

mass of H2O = number of mol * molar mass

= 0.6583*18.02

= 11.86 g

Answer: 11.9 g