Chemistry, 19.03.2020 21:06, angeldawnfick
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.
Answers: 2
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A precipitate of zinc hydroxide can be formed using the reaction below.
ZnCl2(aq) + 2 KO...
ZnCl2(aq) + 2 KO...
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