Question

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 72. g of hydrobromic acid is mixed with 62.3 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 HBr,
MM = 1*MM(H) + 1*MM(Br)
= 1*1.008 + 1*79.9
= 80.908 g/mol


mass(HBr)= 72.0 g

use:
number of mol of HBr,
n = mass of HBr/molar mass of HBr
=(72 g)/(80.91 g/mol)
= 0.8899 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)= 62.3 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(62.3 g)/(40 g/mol)
= 1.558 mol
Balanced chemical equation is:
HBr + NaOH ---> H2O + NaBr


1 mol of HBr reacts with 1 mol of NaOH
for 0.8899 mol of HBr, 0.8899 mol of NaOH is required
But we have 1.558 mol of NaOH

so, HBr is limiting reagent
we will use HBr 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 HBr
= (1/1)*0.8899
= 0.8899 mol


use:
mass of H2O = number of mol * molar mass
= 0.8899*18.02
= 16.03 g
Answer: 16 g