In the reaction MgCl2 + 2KOH → Mg(OH)2 + 2KCI, how many grams of KOH react with 1 mole of MgCl2?
A. 95 g
B. 56 g
C. (1/2)(56 g)
O D. (2)(56 g)

The correct option is;

D. (2)(56 g)

Explanation:

MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl

From the balanced chemical reaction equation, we have;

One mole of MgCl₂ reacts with  two moles of KOH to produce one mole of Mg(OH)₂ and 2 moles of KCl

Therefore, the number of moles of KOH that react with one mole of KCl = 2 moles

The mass, m, of the two moles of KOH = Number of moles of KOH × Molar mass of KOH

The molar mass of KOH = 56.1056 g/mol

∴ The mass, m, of the two moles of KOH = 2 moles × 56.1056 g/mol = 112.2112 grams

The amount in grams of KOH that react with one mole of MgCl₂ = 112.2112 grams ≈ 112 grams =  (2)(56 g).

it's filtration, but this is an assumption so my apologies if i got this