CALCULATE THE MASS OF Na3PO3 PRODUCED AFTER 28.0g OF H3PO3 IS TREATED WITH 42.0g OF NaOH, AND ALSO CALCULATE THE MASS REMAINING OF THE EXCESS REAGENT. H3PO3+ 3NaOH = Na3PO3+ 3H2O

To calculate the mass of Na3PO3 produced, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed, and it dictates the amount of product formed.

1. Start by finding the number of moles of each reactant:
- Moles of H3PO3 = mass of H3PO3 / molar mass of H3PO3
= 28.0 g / (1 x 3 x 16.0 g/mol + 1 x 1 x 1.0 g/mol)
= 0.56 mol

- Moles of NaOH = mass of NaOH / molar mass of NaOH
= 42.0 g / (1 x 22.99 g/mol + 1 x 16.0 g/mol + 1 x 1.0 g/mol)
= 1.08 mol

2. Determine the molar ratio between H3PO3 and Na3PO3 from the balanced equation:
- From the equation: 1 mol H3PO3 reacts with 1 mol Na3PO3
- Thus, the stoichiometry is 1:1.

3. Calculate the moles of Na3PO3 that can be produced based on the limiting reagent:
- Since the stoichiometry is 1:1, the number of moles of Na3PO3 formed would be the same as H3PO3.
Therefore, moles of Na3PO3 produced = 0.56 mol

4. Calculate the mass of Na3PO3 produced:
- Mass of Na3PO3 = moles of Na3PO3 produced × molar mass of Na3PO3
= 0.56 mol × (3 x 22.99 g/mol + 1 x 30.97 g/mol)
= 97.77 g (rounded to two decimal places)

To calculate the mass remaining of the excess reagent, we need to determine the excess reagent first.

5. Determine the molar ratio between NaOH and Na3PO3 from the balanced equation:
- From the equation: 3 mol NaOH reacts with 1 mol Na3PO3
- Thus, the stoichiometry is 3:1.

6. Calculate the moles of Na3PO3 needed to react with the given moles of NaOH:
- Moles of Na3PO3 needed = moles of NaOH × (1 mol Na3PO3 / 3 mol NaOH)
= 1.08 mol × (1 mol Na3PO3 / 3 mol NaOH)
= 0.36 mol

7. Calculate the moles of NaOH remaining:
- Moles of NaOH remaining = moles of NaOH initially - moles of Na3PO3 needed
= 1.08 mol - 0.36 mol
= 0.72 mol

8. Calculate the mass of NaOH remaining:
- Mass of NaOH remaining = moles of NaOH remaining × molar mass of NaOH
= 0.72 mol × (1 x 22.99 g/mol + 1 x 16.0 g/mol + 1 x 1.0 g/mol)
= 34.27 g (rounded to two decimal places)

Therefore, the mass of Na3PO3 produced is 97.77 g, and the mass of NaOH remaining is 34.27 g.