Questions: Uniting reactants Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 42.9 g of hydrobromic acid is mixed with 28 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.

Solution Steps

The balanced chemical equation for the reaction between hydrobromic acid (HBr) and sodium hydroxide (NaOH) is:

$$\mathrm{HBr} + \mathrm{NaOH} \rightarrow \mathrm{NaBr} + \mathrm{H_2O}$$

Calculate the molar masses of the reactants and products:

• Molar mass of HBr: $1.0079 + 79.904 = 80.9119 \, \text{g/mol}$
• Molar mass of NaOH: $22.9898 + 15.9994 + 1.0079 = 39.9971 \, \text{g/mol}$
• Molar mass of NaBr: $22.9898 + 79.904 = 102.8938 \, \text{g/mol}$

First, convert the masses of the reactants to moles:

• Moles of HBr: $\frac{42.9 \, \text{g}}{80.9119 \, \text{g/mol}} = 0.5303 \, \text{mol}$
• Moles of NaOH: $\frac{28.0 \, \text{g}}{39.9971 \, \text{g/mol}} = 0.7001 \, \text{mol}$

Since the reaction ratio is 1:1, the limiting reactant is HBr because it has fewer moles.

Using the moles of the limiting reactant (HBr), calculate the moles of NaBr produced:

• Moles of NaBr: $0.5303 \, \text{mol}$

Convert the moles of NaBr to grams:

• Mass of NaBr: $0.5303 \, \text{mol} \times 102.8938 \, \text{g/mol} = 54.60 \, \text{g}$

Final Answer