Question

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 59. g of hydrobromic acid is mixed with 48.8 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

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)= 48.8 g

number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(48.8 g)/(39.998 g/mol)
= 1.22 mol

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


mass(HBr)= 59.0 g

number of mol of HBr,
n = mass of HBr/molar mass of HBr
=(59.0 g)/(80.908 g/mol)
= 0.7292 mol
Balanced chemical equation is:
NaOH + HBr ---> NaBr + H2O


1 mol of NaOH reacts with 1 mol of HBr
for 1.2201 mol of NaOH, 1.2201 mol of HBr is required
But we have 0.7292 mol of HBr

so, HBr is limiting reagent

So, there will not be any HBr left

Answer: 0.0 g