Question

can anyone help balencing equations in the comments

Answer 1

hold on i meesed up

Answer 2

@Michele_Laino

Answer 3

is your reaction, like this:
\[\Large {\text{Zn + H}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}}{{\text{O}}_{\text{2}}} \to {\text{Zn}}\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\]

Answer 4

\[Zn+HC _{4}H _{3}O _{2}\rightarrow ZN(cH _{3}Coo)_{2}+H _{2}\]

Answer 5

so no H after Coo

Answer 6

sorry, I understand:
\[\Large {\text{Zn + H}}{{\text{C}}_{\text{4}}}{{\text{H}}_{\text{3}}}{{\text{O}}_{\text{2}}} \to {\text{Zn}}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}} \right)_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}\]

Answer 7

ok yes that is it

Answer 8

please wait I'm working on your question...

Answer 9

ok thanks

Answer 10

I'm very sorry, I don't know how to balance such chemical reaction

Answer 11

@chmvijay please can you help here?

Answer 12

@jigglypuff314 please can you help here?

Answer 13

to my knowledge on how to balance equations, I don't believe it is possible...

on the left you have
1 Zn
4 H
4 C
2 O

and on the right you have
1 Zn
8 H
4 C
4 O

because of how they are connected to each other, no matter what coefficients you try to add, it won't even out :-/

I can't help, sorry.

Answer 14

wait... please recheck that is the correct equation?
\[\Large {\text{Zn + H}}{{\text{C}}_{\text{4}}}{{\text{H}}_{\text{3}}}{{\text{O}}_{\text{2}}} \to {\text{Zn}}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}} \right)_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}\]

Answer 15

wait it is not that first 4 is suppose to be a 2