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? The balanced reaction is:. MgCl2 + 2KOH Mg(OH)2 + 2KCl

Answer 1

Answer: The mass of KOH will be 448 grams.

Explanation:

For the given chemical reaction:

$MgCl_2+2KOH\rightarrow Mg(OH)_2+2KCl$

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 = $\frac{2}{1}\times 4=8mol$ of potassium hydroxide.

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Moles of potassium hydroxide = 8 mol

Molar mass of potassium hydroxide = 56 g/mol

Putting values in above equation, we get:

$8mol=\frac{\text{Mass of KOH}}{56g/mol}\\\\\text{Mass of KOH}=448g$

Hence, the mass of KOH will be 448 grams.

Answer 2

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!