Question

What mass of sodium hydroxide, NaOH, would be required to produce 16 g of antacid milk of magnesia?

Answer 1

Approximately 21.96 g of sodium hydroxide (NaOH) would be needed to produce 16 g of milk of magnesia, using the stoichiometry of the reaction MgCl₂ (aq) + 2NaOH(aq) → Mg(OH)₂ (s) + 2NaCl(aq).

Step-by-step explanation:

To determine the mass of sodium hydroxide (NaOH) required to produce 16 g of milk of magnesia, magnesium hydroxide (Mg(OH)₂), we will follow stoichiometry and molar mass calculations based on the chemical reaction between magnesium chloride (MgCl₂) and sodium hydroxide.

Step-by-Step Solution

1. Write down the balanced chemical equation.
   MgCl₂ (aq) + 2NaOH(aq) → Mg(OH)₂ (s) + 2NaCl(aq)
2. Calculate the molar mass of Mg(OH)₂, which is 58.3 g/mol as given.
3. Convert the mass of Mg(OH)₂ to moles using its molar mass.
   16 g Mg(OH)₂ * (1 mol Mg(OH)₂ / 58.3 g) = 0.2745 moles Mg(OH)₂
4. Use the stoichiometry of the balanced reaction to find the moles of NaOH needed. For every mole of Mg(OH)₂ produced, 2 moles of NaOH are required.
   0.2745 moles Mg(OH)₂ * (2 moles NaOH / 1 mole Mg(OH)₂) = 0.549 moles NaOH
5. Calculate the molar mass of NaOH, which is 40.00 g/mol.
6. Convert the moles of NaOH to grams using its molar mass.
   0.549 moles NaOH * (40.00 g NaOH / 1 mole NaOH) = 21.96 g NaOH

Therefore, 21.96 g of sodium hydroxide would be required to produce 16 g of milk of magnesia.