What theoretical mass of NaCl would result from reacting 3.00 moles of NaHCO3 with excess HCl?

The balanced chemical equation for the reaction between NaHCO3 and HCl is:

NaHCO3 + HCl -> NaCl + H2O + CO2

From the equation, we can see that 1 mole of NaHCO3 reacts with 1 mole of HCl to produce 1 mole of NaCl.

Therefore, if 3.00 moles of NaHCO3 react with excess HCl, the theoretical mass of NaCl produced would be 3.00 moles.

To calculate the mass of NaCl, we need to know the molar mass of NaCl, which is 58.44 g/mol.

Mass of NaCl = 3.00 moles x 58.44 g/mol = 175.32 g

Therefore, the theoretical mass of NaCl produced from reacting 3.00 moles of NaHCO3 with excess HCl would be 175.32 grams.

Balanced Chemical Equation:

```
NaHCO3 + HCl → NaCl + H2O + CO2
```

Step 1: Determine the mole ratio between NaHCO3 and NaCl

From the balanced equation, the mole ratio is 1:1. This means that for every 1 mole of NaHCO3 that reacts, 1 mole of NaCl is produced.

Step 2: Calculate the theoretical moles of NaCl

Multiply the given moles of NaHCO3 by the mole ratio to determine the theoretical moles of NaCl:

```
Theoretical moles of NaCl = 3.00 moles NaHCO3 × (1 mole NaCl / 1 mole NaHCO3)
Theoretical moles of NaCl = 3.00 moles
```

Step 3: Calculate the theoretical mass of NaCl

Multiply the theoretical moles of NaCl by its molar mass (58.44 g/mol):

```
Theoretical mass of NaCl = 3.00 moles NaCl × 58.44 g/mol
Theoretical mass of NaCl = 175.32 g
```

Therefore, the theoretical mass of NaCl that would result from reacting 3.00 moles of NaHCO3 with excess HCl is 175.32 g.