Question

In: Chemistry

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride...

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.

Solutions

Expert Solution


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: 0.074 g


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