A precipitate of zinc hydroxide can be formed using the reaction below.

ZnCl2(aq) + 2 KOH(aq) → Zn(OH)2(s) + 2 KCl(aq)

If excess KOH is used, the zinc hydroxide will form zincate ions and dissolve in the solution according to the chemical equation below.

Zn(OH)2(s) + 2 OH-(aq) → Zn(OH)42-(aq)

Which of the following best describes the reaction that occurs when 0.36 mole of ZnCl2 in solution is mixed with 0.54 mole of KOH in solution?
A
KOH is the limiting reagent, and 0.54 mole of Zn(OH)2 precipitate is produced.

B
KOH is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

C
ZnCl2 is the limiting reagent, and no Zn(OH)2 precipitate is produced.

D
ZnCl2 is the limiting reagent, and 0.27 mole of Zn(OH)2 precipitate is produced.

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

може ли да зареждате повече информация, моля