Question

Aqueous hydrobromic acid HBr reacts with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O . What is the theoretical yield of water formed from the reaction of 51.0g of hydrobromic acid and 20.4g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it.

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)= 51.0 g

number of mol of HBr,

n = mass of HBr/molar mass of HBr

=(51.0 g)/(80.908 g/mol)

= 0.6303 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)= 20.4 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(20.4 g)/(39.998 g/mol)

= 0.51 mol

Balanced chemical equation is:

HBr + NaOH ---> H2O + NaBr

1 mol of HBr reacts with 1 mol of NaOH

for 0.630346 mol of HBr, 0.630346 mol of NaOH is required

But we have 0.510026 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)*0.510026

= 0.510026 mol

mass of H2O = number of mol * molar mass

= 0.51*18.02

= 9.19 g

Answer: 9.19 g