Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (Naße) and liquid water (H₂O). Suppose 6.5 g of hydrobromic acid is mixed with 2.30 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.
The Correct Answer and Explanation is :
The maximum mass of water that can be produced from the reaction is approximately 1.036 grams.
Explanation
Step 1: Understanding the Chemical Reaction
The balanced chemical equation for the reaction between hydrobromic acid (HBr) and sodium hydroxide (NaOH) is:
[ \text{HBr (aq)} + \text{NaOH (s)} \rightarrow \text{NaBr (aq)} + \text{H}_2\text{O (l)} ]
This equation indicates that one mole of HBr reacts with one mole of NaOH to produce one mole of sodium bromide (NaBr) and one mole of water (H₂O).
Step 2: Calculating Molar Masses
- Molar mass of HBr = 80.912 g/mol
- Molar mass of NaOH = 39.997 g/mol
- Molar mass of H₂O = 18.015 g/mol
Step 3: Determining Moles of Reactants
- Moles of HBr = Mass / Molar mass = ( \frac{6.5 \text{ g}}{80.912 \text{ g/mol}} = 0.0803 \text{ mol} )
- Moles of NaOH = Mass / Molar mass = ( \frac{2.30 \text{ g}}{39.997 \text{ g/mol}} = 0.0575 \text{ mol} )
Step 4: Identifying the Limiting Reactant
The reaction is a 1:1 ratio between HBr and NaOH. Since there are fewer moles of NaOH compared to HBr, NaOH is the limiting reactant.
Step 5: Calculating Maximum Mass of Water
Given the limiting reactant (NaOH) dictates the amount of products formed:
- Moles of water produced = Moles of NaOH = 0.0575 mol
- Mass of water = Moles of water × Molar mass of H₂O = ( 0.0575 \text{ mol} \times 18.015 \text{ g/mol} = 1.036 \text{ g} )
Thus, the reaction between 6.5 grams of hydrobromic acid and 2.30 grams of sodium hydroxide can produce up to 1.036 grams of water under ideal conditions. This calculation assumes complete reaction of the limiting reactant with no side reactions or losses.