Question

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O . Suppose 4.9 g of hydrobromic acid is mixed with 3.86 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Round your answer to 2 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)= 4.9 g

number of mol of HBr,

n = mass of HBr/molar mass of HBr

=(4.9 g)/(80.908 g/mol)

= 6.056*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)= 3.86 g

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(3.86 g)/(39.998 g/mol)

= 9.65*10^-2 mol

Balanced chemical equation is:

HBr + NaOH ---> NaBr + H2O

1 mol of HBr reacts with 1 mol of NaOH

for 0.060563 mol of HBr, 0.060563 mol of NaOH is required

But we have 0.096505 mol of NaOH

so, HBr is limiting reagent

we will use HBr 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 HBr

= (1/1)*0.060563

= 0.060563 mol

mass of NaBr = number of mol * molar mass

= 6.056*10^-2*1.029*10^2

= 6.2 g

Answer: 6.2 g