Question

Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . Suppose 17. g of hydrobromic acid is mixed with 2.44 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: The mass of NaBr that can be produced is 6.3 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For HBr:

Given mass of HBr = 17 g

Molar mass of HBr = 81 g/mol

Putting values in equation 1, we get:

$\text{Moles of HBr}=\frac{17g}{81g/mol}=0.210mol$

• For NaOH:

Given mass of NaOH = 2.44 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{2.44g}{40g/mol}=0.061mol$

The chemical equation for the reaction of HBr and NaOH follows:

$HBr+NaOH\rightarrow NaBr+H_2O$

By Stoichiometry of the reaction:

1 mole of NaOH reacts with 1 mole of HBr

So, 0.061 moles of NaOH will react with = $\frac{1}{1}\times 0.061=0.061mol$ of HBr

As, given amount of HBr is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of NaOH produces 1 mole of NaBr

So, 0.061 moles of NaOH will produce = $\frac{1}{1}\times 0.061=0.061moles$ of carbon dioxide

Now, calculating the mass of NaBr from equation 1, we get:

Molar mass of NaBr = 103 g/mol

Moles of NaBr = 0.061 moles

Putting values in equation 1, we get:

$0.061mol=\frac{\text{Mass of NaBr}}{103g/mol}\\\\\text{Mass of NaBr}=(0.061mol\times 103g/mol)=6.28g$

Hence, the mass of NaBr that can be produced is 6.3 grams