Question

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 2.4 g of hydrobromic acid is mixed with 0.876 g of sodium hydroxide. Calculate the maximum mass of sodium bromide 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)= 2.4 g

use:

number of mol of HBr,

n = mass of HBr/molar mass of HBr

=(2.4 g)/(80.91 g/mol)

= 2.966*10^-2 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)= 0.876 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(0.876 g)/(40 g/mol)

= 2.19*10^-2 mol

Balanced chemical equation is:

HBr + NaOH ---> NaBr + H2O

1 mol of HBr reacts with 1 mol of NaOH

for 2.966*10^-2 mol of HBr, 2.966*10^-2 mol of NaOH is required

But we have 2.19*10^-2 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of NaBr,

MM = 1*MM(Na) + 1*MM(Br)

= 1*22.99 + 1*79.9

= 102.89 g/mol

According to balanced equation

mol of NaBr formed = (1/1)* moles of NaOH

= (1/1)*2.19*10^-2

= 2.19*10^-2 mol

use:

mass of NaBr = number of mol * molar mass

= 2.19*10^-2*1.029*10^2

= 2.253 g

Answer: 2.25 g