Question

Aqueous HBr reacts with solid NaOH to produce sodium bromide NaBr and liquid water. What is the theoretical yield of water formed from the reaction of 34.0 g of HBr and 6.05g of NaOH? 3 sig figs please.

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Answer 1

The reaction equation is as follows:

HBr (aq) + NaOH(s) ----> NaBr + H2O (l)

Number of moles of HBr = mass/ molar mass = 34/ 80.91 = 0.4202 mol

Number of moles of NaOH = 6.05/40 = 0.151 mol

From the reaction equation, 1 mole of HBr reacts with one mole of NaOH. Here, 0.151 mol of NaOH is only present. NaOH is the limiting reagent.

Hence, 0.151 mol of NaOH reacts with 0.151 mol of HBr to give 0.151 mol of water.

Mass of water = number of moles*molar mass = 0.151*18 = 2.7225 g

Theoretical yield of water = 2.7225 g