Question

Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H0). What is the theoretical yield of water formed from the reaction of 23.5 g of hydrobromic acid and 17.2, of sodium hydroxide? Round your answer to 3 significant figures.

Answer 1

Solution HBr (aq) + NaOH(s) - Imole i mole NaBraq) + H2O(l) Imole Malarmass of HBr = 1 + 80 = 81 g/mole Molarmass of NaOH = 23+ 16+ / B = 40g/mole so Imole HBr = 8lg. and Imole NaOH = nog From the reaction it is clear the 1 81g HBr reacts with yog NaOH ,, 23.59 HBr reacts with 40g x 23.5g NaOH 81 8 = 116 g NaOH so we can say that HBr is limiting reagent and we can calculate theoretical yield of water either by - 23.5g HBr or 1.6g. NaOH. Molar mass of H20 = 2X1 +16 B = 18 g/mole means Scanned wimol CamScanner

I dig HBr form 1. 23.5g HBr form 18 g H₂O log X23.59 Hoo Olg = 5.229 H2O Ans:- Theoretical yield of water formed is 5.22 g LONNELLE CS Scanned with CamScanner