Answer:
Option B is correct. KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.
Explanation:
Step 1: Data given
Number of moles ZnCl2 = 0.36 moles
Number of moles KOH = 0.54 moles
Step 2: The balanced equations
ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)
For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl
Step 3: Calculate the limiting reactant.
KOH is the limiting reactant. It will completely be consumed (0.54 moles). ZnCl2 is in excess. There will react 0.54/2 = 0.27 moles
There will remain 0.36 - 0.27 = 0.09 moles.
Step 4: The products
There will be produced 2 moles KCl and 1 mol Zn(OH)2. Zn(OH)2 is the precipitate produced.
For 1 mol ZnCl2 we need 2 moles KOH to produce 1 mol Zn(OH)2 and 2 moles KCl
For 0.54 moles KOH, we will produce 0.27 moles precipitate (Zn(OH)2)
Option A is not correct because 0.27 mol of Zn(OH)2 will precipitate, not 0.54 mol
Option B is correct
Option C is not correct because ZnCl2 is not the limiting reactant, but the excess reactant
Option D is not correct because ZnCl2 is not the limiting reactant, but the excess reactant
