Which equation shows how to calculate how many grams (g) of KOH would be needed to fully react with 4 mol Mg(OH)2? The balanced reaction is:. MgCl2 + 2KOH Mg(OH)2 + 2KCl

Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)

moles KOH = 4 x 2 / 1

= 8 moles of KOH

molar mass KOH = 56 g/mol

mass of KOH = n x mm

mass of KOH = 8 x 56

= 448 g of KOH

hope this helps!

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RomeliaThurston RomeliaThurston · Answer 2 · 2019-07-03T09:30:05+02:00

Answer: The mass of KOH will be 448 grams.

Explanation:

For the given chemical reaction:

[tex]MgCl_2+2KOH\rightarrow Mg(OH)_2+2KCl[/tex]

By Stoichiometry of the reaction:

1 mole of magnesium chloride reacts with 2 moles of potassium hydroxide.

So, 4 moles of magnesium chloride will react with = [tex]\frac{2}{1}\times 4=8mol[/tex] of potassium hydroxide.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of potassium hydroxide = 8 mol

Molar mass of potassium hydroxide = 56 g/mol

Putting values in above equation, we get:

[tex]8mol=\frac{\text{Mass of KOH}}{56g/mol}\\\\\text{Mass of KOH}=448g[/tex]

Hence, the mass of KOH will be 448 grams.