Question

Aqueous hydrochloric acid

HCl

will react with solid sodium hydroxide

NaOH

to produce aqueous sodium chloride

NaCl

and liquid water

H2O

. Suppose 0.36 g of hydrochloric acid is mixed with 0.164 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)= 0.36 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(0.36 g)/(36.458 g/mol)
= 9.874*10^-3 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)= 0.164 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.164 g)/(39.998 g/mol)
= 4.1*10^-3 mol
Balanced chemical equation is:
HCl + NaOH ---> H2O + NaCl


1 mol of HCl reacts with 1 mol of NaOH
for 9.874*10^-3 mol of HCl, 9.874*10^-3 mol of NaOH is required
But we have 4.1*10^-3 mol of NaOH

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)*4.1*10^-3
= 4.1*10^-3 mol


use:
mass of H2O = number of mol * molar mass
= 4.1*10^-3*18.02
= 7.4*10^-2 g
Answer: 7.4*10^-2 g