1. Write the balanced chemical equation:

The reaction between methanol (CH3OH) and ammonia (NH3) produces methylamine (CH3NH2) and water (H2O).

CH3OH + NH3 → CH3NH2 + H2O

2. Calculate the molar mass of each reactant:

* CH3OH (methanol): 12.01 g/mol (C) + 4.03 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol

* NH3 (ammonia): 14.01 g/mol (N) + 3.02 g/mol (H) = 17.03 g/mol

3. Convert the mass of methanol to moles:

* Moles of CH3OH = (mass of CH3OH) / (molar mass of CH3OH)

* Moles of CH3OH = 21 g / 32.04 g/mol = 0.655 mol

4. Determine the mole ratio from the balanced equation:

* The balanced equation shows a 1:1 mole ratio between CH3OH and NH3. This means that for every 1 mole of CH3OH, you need 1 mole of NH3.

5. Calculate the moles of NH3 needed:

* Since the mole ratio is 1:1, you need 0.655 mol of NH3.

6. Convert the moles of NH3 to grams:

* Mass of NH3 = (moles of NH3) * (molar mass of NH3)

* Mass of NH3 = 0.655 mol * 17.03 g/mol = 11.15 g

Therefore, you need approximately 11.15 grams of NH3 to react exactly with 21 grams of CH3OH.