Question

Which equation shows how to calculate how many grams (

g) of KOH would be needed to fully react with 4 mol Mg(OH)₂?
Options:

A. (1 mol Mg(OH)₂ / 1) × (2 mol KOH / 1 mol Mg(OH)₂) × (56.10 g KOH / 1 mol KOH)
B. (4 mol Mg(OH)₂ / 1) × (1 mol KOH / 2 mol Mg(OH)₂) × (56.10 g KOH / 1 mol KOH)
C. (4 mol Mg(OH)₂ / 1) × (2 mol KOH / 1 mol Mg(OH)₂) × (56.10 g KOH / 1 mol KOH)
D. (1 mol MgCl₂ / 1) × (2 mol KOH / 1 mol MgCl₂) × (56.10 g KOH / 1 mol KOH)

Answer 1

The correct equation to calculate the mass of KOH needed to react with 4 moles of Mg(OH)2 is Option C, which uses stoichiometry principles to apply the right mole ratio and the molar mass of KOH for the conversion.

Step-by-step explanation:

The question is about calculating the mass of potassium hydroxide (KOH) needed to fully react with magnesium hydroxide (Mg(OH)2) using stoichiometry. To find the mass of KOH required, we first need the mole ratio of Mg(OH)2 to KOH from the balanced chemical equation, and then use the molar mass of KOH to convert from moles to grams.

Looking at the options:

• Option A is incorrect because it is not based on the initial 4 moles of Mg(OH)2.
• Option B reverses the mole ratio of KOH to Mg(OH)2.
• Option C correctly represents the situation: (4 mol Mg(OH)2 / 1) × (2 mol KOH / 1 mol Mg(OH)2) × (56.10 g KOH / 1 mol KOH).
• Option D is incorrect because it refers to MgCl2, which is not involved in this scenario.

Therefore, the correct equation to use for determining the amount of KOH needed is Option C.