Question

Aqueous hydrochloric acid (HCI) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H,0). Suppose 13. g of hydrochloric acid is mixed with 8.54 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. 04 0. X 5 ?

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)= 13.0 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(13 g)/(36.46 g/mol)

= 0.3566 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)= 8.54 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(8.54 g)/(40 g/mol)

= 0.2135 mol

Balanced chemical equation is:

HCl + NaOH ---> H2O + NaCl

1 mol of HCl reacts with 1 mol of NaOH

for 0.3566 mol of HCl, 0.3566 mol of NaOH is required

But we have 0.2135 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)*0.2135

= 0.2135 mol

use:

mass of H2O = number of mol * molar mass

= 0.2135*18.02

= 3.847 g

Answer: 3.8 g