calculate the mass of NaOH needed for complete neutralisation of 73g of dil HCl and also calculate the mass of NaCl obtained as a product.

To calculate the mass of NaOH needed for the complete neutralization of 73 g of dilute HCl and the mass of NaCl obtained, follow these steps:

1. Write the balanced chemical equation for the reaction:

\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

2. Calculate the moles of HCl:

- Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol

- Moles of HCl = \(\frac{\text{mass of HCl}}{\text{molar mass of HCl}}\)

\[ \text{Moles of HCl} = \frac{73 \text{ g}}{36.5 \text{ g/mol}} \approx 2 \text{ mol} \]

3. Determine the moles of NaOH needed:

According to the balanced equation, 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the moles of NaOH required = moles of HCl.

- Moles of NaOH = 2 mol

4. Calculate the mass of NaOH needed:

- Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol

- Mass of NaOH = moles of NaOH × molar mass of NaOH

\[ \text{Mass of NaOH} = 2 \text{ mol} \times 40 \text{ g/mol} = 80 \text{ g} \]

5. Calculate the mass of NaCl formed:

According to the balanced equation, 1 mole of NaOH produces 1 mole of NaCl.

- Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol

- Mass of NaCl = moles of NaCl × molar mass of NaCl

\[ \text{Mass of NaCl} = 2 \text{ mol} \times 58.5 \text{ g/mol} = 117 \text{ g} \]

Summary:

- Mass of NaOH needed: 80 g

- Mass of NaCl formed: 117 g