Question

Aqueous hydrobromic acid hbr will react with solid sodium hydroxide naoh to produce aqueous sodium bromide nabr and liquid water h2o . suppose 20.2 g of hydrobromic acid is mixed with 5.7 g of sodium hydroxide. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to 2 significant digits.

Answer 1

 Balanced equation is

HBr + NaOH ----> NaBr + H2O

Using molar masses

80.912 g HBr reacts with  39.997 g of Naoh to give 18.007 g water

so 1 gram of NaOH reacts with 2.023 g of HBR   
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction

Mass  of water produced  =    (5.7 * 18.007 / 39.997  =  2.6 g to 2 sig figs

Answer 2

Answer:

The maximum mass of water that could be produced by the chemical reaction is 2.6 grams.

Explanation:

$HBr(aq)+NaOH(s)\rightarrow NaBr(aq)+H_2O(l)$

Moles of HBr = $\frac{20.2 g}{81 g/mol}=0.2494 mol$

Moles of sodium NaOH = $\frac{5.7 g}{40 g/mol}=0.1425 mol$

According to reaction 1 mol of HBr is reacting with 1 mol of NaOH.

Then 0.2494 moles of HBr will react with :

$\frac{1}{1}\times 0.2494 mol=0.2494 mol$ of NaOH.

According to reaction 1 mol of NaOH is reacting with 1 mol of HBr.

Then 0.1425 moles of sodium hydroxide will react with :

$\frac{1}{1}\times 0.1425 mol=0.1425 mol$ of HBr.

As we can see that NaOH is in limiting amount of moles. So, the amount of water formed will depend upon the amount of NaOH.

According to reaction, 1 mole NaOH gives 1 mole of water.

Then 0.1425 moles of NaOH will give:

$\frac{1}{1}\times 0.1425 mol=0.1425 mol$ of water.

Mass of the 0.1425 mol of water:

0.1425 mol × 18 g/mol = 2.565 g ≈ 2.6 g.

The maximum mass of water that could be produced by the chemical reaction is 2.6 grams.