How do I find the net ionic equations for these

I would appreciate a painstaking explanation as well as step-guidance. Thank you all.

What are the net ionic equations for the following reactions:

AgNO_3 (aq) with Ba(OH)_2 (aq).

HCN (aq) with COH (aq) --> H_2O (L) + KCN (aq).

Once more, thanks.

To get net ionic, get balanced equation, ionic equation, then cancel spectator ions and get net ionic

Number 1

AgNO3 (aq) + Ba(OH)2 (aq)

This is a double replacement reaction, since both reactants consist of two "parts". The products will just have the metal ion switched.

Step 1 writing the full equation

2AgNO3 (aq) + Ba(OH)2 (aq) yields Ba(NO3)2 (aq) and 2AgOH (s)

I shall explain.

First, like I said, it's double replacement so switch the metals like i did.

Then, put the right subscript according to charge. The barium ion is Ba2+ since it bonded with 2 hydroxides, which are OH1-, so pair the Ba with 2 NO3s, since the charge of a NO3 ion is -1. Ag is always +1, and as previously mentioned, OH is -1 so AgOH is fine.

Then, if you know your solubility rules (memorize these, you will need them http://bit.ly/od9NfZ), you will know that Silver hydroxide is insoluble.

Then, balance the equation. I think you know how to do this? Based on the law of conservation of mass, the number of atoms of one element should be the same on both sides. So add coefficients such that this law is satisfied. There are 2 NO3s, 2 Ag's, 1 Ba, and 2 OH's on both sides so this is balanced.

Step 2 (ionic equation)

(2Ag+) (aq) + (2(NO3)-) (aq) + (Ba2+) (aq) + (2OH-) (aq) yields (Ba2+) (aq) + (2(NO3)-) (aq) + 2AgOH (s)

What I did:

Split up compounds that are soluble (aq) leave the rest

When there are coefficients, apply each coefficient to each ion in the compound. It was 2AgNO3, so when split up, it becomes (2Ag+) (aq) + (2(NO3)-) (aq)

Write the charges too. For example, even though there are two NO3s and the total charge of two NO3s is -2, write it as 2NO3- instead of (2NO3)2-