Queston 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 + 2KCI
OA. mol Mg(OH),
2 mol KOH
mol Mg(0H),
56.10 g KOH
1 mol KOH
2 mol KOH
1 mol Mg Cl,
56.10 g KOH
1mol KOH
OB. mol Mgch
1
1mol Mg(OH),
2 mol KOH
1mol Ma(0H),
56.10 g KOH
1 molkOH
mol Mg (OH,
1 mol KOH
2 mol Ma(OH)
56 10 g KOH
1mol KOH
OD.

C

Explanation:

We know that the molar mass of KOH is 56 gmol-1

Molar mass of Mg(OH)2 is 58.31 g/mol

Now; the balanced reaction equation is; MgCl2 + 2KOH → Mg(OH)2 + 2KCI

Hence;

2 moles of KOH yields 1 mole of Mg(OH)2

4 moles of KOH yields;

4 moles of KOH/1 mol * 1   mole Mg(OH)2 /2 moles KOH * 58.31 g Mg(OH)2 / 1 mole Mg(OH)2

So option c is correct

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