Question

Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and lic the theoretical yield of water formed from the reaction of 1.6 g of hydrobromic acid and 1.3 g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it liquid water (H20). what is

Answer 1

The above reaction proceeds as :
HBr + NaOH ---> NaBr + H₂O

We use the mole concept to analyse the problem.

As we can see, 1 mole of HBr reacts with 1 mole of NaOH to produce 1 mole of water.

That is, every 81g of HBr reacts with 40 g of NaOH to produce 18 g of water.

We are provided with 1.6g and 1.3 g of HBr and NaOH respectively.

We first find the limiting reagent.

As 81 g HBr reacts with 40 g NaOH, 1.3 g NaOH should react with : (81 / 40 ) * 1.3 = 2.6325 g of HBr. However, we have 1.6 g of HBr, which is therefore in short supply.  

Therefore, HBr is the limiting reagent in this reaction.

1.6 g of HBr reacts with NaOH to produce = ( 18 / 81) * 1.6 = 0.356 g water