Final answer:
To produce 16 g of magnesium hydroxide (milk of magnesia) from the reaction with sodium hydroxide, we first calculate the molar masses of both compounds. Next, we use stoichiometry to determine the moles of sodium hydroxide needed, and finally convert these moles to grams to find that approximately 22 g of sodium hydroxide is required.
Step-by-step explanation:
The question asks how much sodium hydroxide (NaOH) would be required to produce 16 g of magnesium hydroxide (Mg(OH)₂), also known as milk of magnesia, by the reaction between magnesium chloride (MgCl₂) and sodium hydroxide (NaOH). The balanced chemical equation for this reaction is:
MgCl₂(aq) + 2NaOH(aq) → Mg(OH)₂(s) + 2NaCl(aq)
First, we need to calculate the molar masses of the compounds involved:
- Mg(OH)₂: 24.305 (Mg) + 2 × 15.999 (O) + 2 × 1.008 (H) = 58.3195 g/mol
- NaOH: 22.990 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol
Using the molar mass of Mg(OH)₂, we determine how many moles of Mg(OH)₂ correspond to the given 16 g:
16 g Mg(OH)₂ × (1 mol / 58.3195 g) = 0.2744 mol Mg(OH)₂
Since the reaction requires 2 moles of NaOH for every mole of Mg(OH)₂ produced, we can calculate the moles of NaOH needed:
0.2744 mol Mg(OH)₂ × (2 mol NaOH / 1 mol Mg(OH)₂) = 0.5488 mol NaOH
Finally, we convert moles of NaOH to grams using the molar mass of NaOH:
0.5488 mol NaOH × (39.997 g/mol NaOH) = 21.951 g NaOH
Therefore, the mass of sodium hydroxide required is approximately 22 g, which corresponds to option (a).