How do I balance this reaction when producing methane using the reactants:

$$\ce{CH3COONa(s) + NaOH(s) + Ca(OH)2 -> ?}$$

The products are partly

$\ce{CH4(g) + Na2CO3(s)}$

What else is missing in the products? Furthermore, how do I balance this?

Thanks in advance.


closed as off-topic by M.A.R. ಠ_ಠ, Wildcat, ron, airhuff, paracetamol May 6 '17 at 16:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is a homework question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. $\endgroup$ – M.A.R. ಠ_ಠ May 6 '17 at 12:02

The answer is:

$$\ce{2CH3COONa(s) + 2NaOH(s) + Ca(OH)2(s) <=> 2CH4(g) + Na2CO3(s) + CaCO3(s) + H2O}$$

You have to get same number of equal atoms in each side of the reaction (this technique is called mass balance). If there were charges, they must be balanced also.


There are $2\times 3 + 2\times 1 + 2\,$ hydrogen atoms in left side, and there are $2\times 4 + 2$ $\ce{H}$ in the right side. So both sides give $10$ hydrogen atoms.

I hope it helps.


Not the answer you're looking for? Browse other questions tagged or ask your own question.