Question

Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 47. g of hydrobromic acid is mixed with 9.51 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over 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)= 47.0 g

use:
number of mol of HBr,
n = mass of HBr/molar mass of HBr
=(47 g)/(80.91 g/mol)
= 0.5809 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)= 9.51 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(9.51 g)/(40 g/mol)
= 0.2378 mol
Balanced chemical equation is:
HBr + NaOH ---> NaBr + H2O


1 mol of HBr reacts with 1 mol of NaOH
for 0.5809 mol of HBr, 0.5809 mol of NaOH is required
But we have 0.2378 mol of NaOH

so, NaOH is limiting reagent

According to balanced equation
mol of HBr reacted = (1/1)* moles of NaOH
= (1/1)*0.2378
= 0.2378 mol
mol of HBr remaining = mol initially present - mol reacted
mol of HBr remaining = 0.5809 - 0.2378
mol of HBr remaining = 0.3431 mol


Molar mass of HBr,
MM = 1*MM(H) + 1*MM(Br)
= 1*1.008 + 1*79.9
= 80.908 g/mol

use:
mass of HBr,
m = number of mol * molar mass
= 0.3431 mol * 80.91 g/mol
= 27.76 g
Answer: 28 g