English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Zn+CrCl3--->CrCl2+ZnCl2

2006-10-22 09:48:02 · 5 answers · asked by BWW 3 in Education & Reference Homework Help

5 answers

Is this a chemistry equation? I love chemistry! Okay...

I came out to...
3Zn + 6CrCl3 --> 6CrCl2 + 3ZnCl2

It's right!!! I double checked! As for the people above me... the second guy said 2Zn + 2CrCl3 --> 2CrCl2 + Cl2. First of all, it's wrong because you can't move the Zn over from the right to the left side.

Mine is balanced because there are 3 Zn atoms on each side, 6 Cr atoms on each side, and 18 Cl atoms on each side. Okay... the guy above me and the young lady below me are right. The answer is...

Zn + 2CrCl3 --> 2CrCl2 + ZnCl2... Simple mistake. I'm sorry.

2006-10-22 09:58:19 · answer #1 · answered by randkcarpenterfan 3 · 0 1

The purpose of balancing an equation is to get the same number of atoms of each molecule on each side of the equation. In this case the equation is:
Zn + CrCl3 ---> CrCl2 + ZnCl2

# of atoms:
Zn = 1 Zn= 1
Cr = 1 Cr= 1
Cl=3 Cl= 4

To balance it, insert a 2 before CrCl3 and another 2 before CrCl2 ( The 2 is distributed to both Cr and Cl) It might not always be this easy, sometimes you really have to struggle to find the right numbers by trial and error. Now you see the resulting number of atoms in the balanced equation:

Zn + 2CrCl3 ---> 2CrCl2 + ZnCl2

Zn = 1 Zn = 1
Cr = 2 Cr = 2
Cl = 6 Cl = 6

Hope this helps...good luck ;-)

2006-10-22 10:05:06 · answer #2 · answered by msdrosi 3 · 0 0

Zn+2CrCl3-->2CrCl2+ZnCl2

quite easy...this reaction seems unlikely though

the guy below is okay too, but his need to be reduced...divide through by 3 and we are in line

2006-10-22 09:55:31 · answer #3 · answered by Anonymous · 0 0

Sorry, but wish I knew Trig or Calculus

2006-10-22 09:50:59 · answer #4 · answered by Lucifer Sam 5 · 0 0

2Zn+2CrCl3--->2CrCl2+Cl2
guy above me, this is simple chemistry

2006-10-22 09:55:13 · answer #5 · answered by chubs353 2 · 0 0

fedest.com, questions and answers