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 25.9 g of hydrobromic acid and 19.2 g of sodium hydroxide?

Round to 3 significant digits

Answer 1

Mass of HBr = 25.9 g

Molar mass of HBr = for H + for Br

Molar mass of HBr = 1.00794 + 79.904 g / mol

Molar mass of HBr = 80.91194 g/ mol

Moles of HBr = mass of HBr / Molar mass of HBr

Moles of HBr = 25.9 / 80.91194 mol

Moles of HBr = 0.32010109 mol

Mass of NaOH = 19.2 g

Molar mass of NaOH = 39.997 g/mol

Moles of NaOH = 19.2 / 39.997 mol

Moles of NaOH = 0.480036 mol

Moles of HBr is less than Moles of NaOH,hence HBr is limiting reagent.

HBr +.   NaOH -----> NaBr +. H₂O

At t=0 0.32010109 0.480036 0 0

at time 0 0.480036 - 0.32010109 0.32010109 0.32010109

Hence

Moles of water = 0.32010109 mol

Molar mass of water = 18.01528 g/ mol

Moles = mass / Molar mass

or

Mass of water = Moles of water *Molar mass of water

Mass of water = 0.32010109 *18.01528 g

Mass of water = 5.766710 g

in 3 significant figures

Mass of water = 5.77 g