Aqueous hydrobromic acid will react with solid sodium hydroxide to produce aqueous sodium bromide and liquid water. Suppose 61. g of hydrobromic acid is mixed with 41.1 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction

This is a Limiting Reagent (LR) problem. You know that because amounts are given for BOTH reactants.

1. Write and balance the equation.
2. Convert HBr to mols. mols = grams/molar mass = ?
3. Convert NaOH to mols. mols = grams/molar mass = ?
4. Using the coefficients in the balanced equation, convert mols HBr to mols H2O.
5. Do the same and convert mols NaOH to mols H2O.
6. I is likely that the values for mols H2O in steps 4 and 5 will not agree. The correct answer in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR.
7. Finally, convert mols H2O to grams. grams = mols x molar mass = ?

14g

To determine the maximum mass of water that could be produced, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed and determines the amount of product that can be formed.

Let's find the moles of each reactant:

1. Hydrobromic acid (HBr):
- Mass of HBr: 61.0 g
- Molar mass of HBr: 1.01 g/mol (H) + 79.90 g/mol (Br) = 80.91 g/mol
- Moles of HBr = Mass / Molar mass = 61.0 g / 80.91 g/mol = 0.755 mol

2. Sodium hydroxide (NaOH):
- Mass of NaOH: 41.1 g
- Molar mass of NaOH: 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol
- Moles of NaOH = Mass / Molar mass = 41.1 g / 39.99 g/mol = 1.03 mol

To determine the stoichiometry of the reaction, we compare the coefficients of the balanced equation:

HBr + NaOH -> NaBr + H2O

From the balanced equation, we can see that the ratio of HBr to NaOH is 1:1. So, for complete reaction:
- Moles of HBr = 0.755 mol
- Moles of NaOH = 0.755 mol

Since the mole ratio of HBr to NaOH is 1:1, HBr is the limiting reactant because it will be completely consumed, leaving some excess NaOH.

Now, let's find the moles of water produced:
- Moles of H2O = Moles of HBr = 0.755 mol

Finally, we can calculate the mass of water produced:

1. Molar mass of water (H2O): 1.01 g/mol (H) + 16.00 g/mol (O) = 17.01 g/mol

2. Mass of water produced = Moles of H2O × Molar mass of water
= 0.755 mol × 17.01 g/mol
= 12.86355 g

Therefore, the maximum mass of water that could be produced is approximately 12.9 g.

To find the maximum mass of water that can be produced, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

We'll start by calculating the number of moles of each reactant:

1. Calculate the number of moles of hydrobromic acid (HBr):
- Use the formula: number of moles = mass / molar mass
- The molar mass of hydrobromic acid (HBr) is approximately 80.91 g/mol.
- Therefore, the number of moles of HBr is: 61. g / 80.91 g/mol = 0.754 mol

2. Calculate the number of moles of sodium hydroxide (NaOH):
- Use the same formula: number of moles = mass / molar mass
- The molar mass of sodium hydroxide (NaOH) is approximately 40.00 g/mol.
- Therefore, the number of moles of NaOH is: 41.1 g / 40.00 g/mol = 1.03 mol

Now, we need to determine the stoichiometry of the reaction to find the balanced equation:
- Based on the balanced equation: 2HBr + 2NaOH -> 2NaBr + H2O
- This means that for every 2 moles of HBr, we need 2 moles of NaOH to completely react.

Next, we'll compare the mole ratios of HBr and NaOH. Since the mole ratio is 2:2, which reduces to 1:1, we have the same number of moles for both reactants. Therefore, the reaction is balanced.

Now, to determine the maximum mass of water produced, we need to calculate the number of moles of water produced. From the balanced equation, we see that every 2 moles of HBr produces 1 mole of water.

Since the mole ratio of HBr and water is 2:1, and the number of moles of HBr is 0.754 mol, the maximum number of moles of water that can be produced is 0.754 mol / 2 = 0.377 mol.

Finally, we calculate the maximum mass of water produced using the formula:

mass = number of moles × molar mass

The molar mass of water (H2O) is approximately 18.02 g/mol.

Therefore, the maximum mass of water that could be produced is: 0.377 mol × 18.02 g/mol ≈ 6.8 g.

So, the maximum mass of water produced by the reaction is approximately 6.8 grams.