Question

Aqueous hydrochloric acid (HCI) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water
(H2O). If 0.946 g of water is produced from the reaction of 2.9 g of hydrochloric acid and 4.5 g of sodium hydroxide, calculate the percent yield of water

Answer 1

To calculate the percent yield of water from the reaction, we'll follow these steps:

1. **Write the balanced chemical equation for the reaction:**
\[ \text{HCl}(aq) + \text{NaOH}(s) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l) \]

2. **Find the moles of each reactant:**

- **Molar mass of HCl:**
\[ \text{HCl} = 1.01 \, \text{g/mol (H)} + 35.45 \, \text{g/mol (Cl)} = 36.46 \, \text{g/mol} \]
Moles of HCl:
\[ \text{moles of HCl} = \frac{2.9 \, \text{g}}{36.46 \, \text{g/mol}} = 0.0795 \, \text{mol} \]

- **Molar mass of NaOH:**
\[ \text{NaOH} = 22.99 \, \text{g/mol (Na)} + 15.999 \, \text{g/mol (O)} + 1.01 \, \text{g/mol (H)} = 39.999 \, \text{g/mol} \]
Moles of NaOH:
\[ \text{moles of NaOH} = \frac{4.5 \, \text{g}}{39.999 \, \text{g/mol}} = 0.1125 \, \text{mol} \]

3. **Determine the limiting reactant:**

The balanced reaction shows a 1:1 molar ratio between HCl and NaOH. Comparing the available moles:
- Moles of HCl: 0.0795 mol
- Moles of NaOH: 0.1125 mol

Since HCl has fewer moles, it is the limiting reactant.

4. **Calculate the theoretical yield of water:**

The balanced equation shows that 1 mole of HCl produces 1 mole of water. Therefore, moles of water produced will be equal to moles of the limiting reactant (HCl):
\[ \text{moles of H}_2\text{O} = 0.0795 \, \text{mol} \]

- **Molar mass of water (H\(_2\)O):**
\[ \text{H}_2\text{O} = 2 \times 1.01 \, \text{g/mol (H)} + 15.999 \, \text{g/mol (O)} = 18.02 \, \text{g/mol} \]
Theoretical mass of water:
\[ \text{mass of water} = 0.0795 \, \text{mol} \times 18.02 \, \text{g/mol} = 1.43319 \, \text{g} \]

5. **Calculate the percent yield of water:**
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
Here, the actual yield of water is given as 0.946 g.
\[
\text{Percent Yield} = \left( \frac{0.946 \, \text{g}}{1.43319 \, \text{g}} \right) \times 100 \approx 66.0 \, \%
\]

So, the percent yield of water is approximately 66.0%.