AgNO3 + K2Cr2O7 --> double replacement
AgNO3 + K2Cr2O7 --> AgCr2O7 + K2NO3
The formula for silver dichromate is Ag2Cr2O7 and for potassium nitrate is KNO3. Of course this needs to be balanced again.
Please tell me if I've balanced these right.
Br2 + NiI3 --> single replacement, element is a non-metal
Br2 + NiI3 --> I3 + NiBr2
Br2 + NiI3 ==> I2 + NiBr3 and this needs to be balanced again. What are you doing wrong in these first two? Your aren't paying attention to the valence of the atoms. In the first one, Ag is +1 and dichromate ion is -2. In the second one, I2 is one of those elements that are diatomic in the free state. Remember a few days ago I wrote those as H2, N2, O2, F2, Br2, Cl2, I2, and their cousins P4 and S8. Ni in NiI3 must be +3 (since I is -1) so the formula for nickel(III) bromide is NiBr3.
Cl2 + Mg3N2 --> single replacement, element is a non-metal
Cl2 + Mg3N2 --> N2 + Mg3Cl2
Once again, you are relying on the subscript to write the formula and to help balance (maybe, maybe not) but this should be
Cl2 + Mg3N2 ==> N2 + MgCl2
The equation I'm stuck on is....
HCl + Mo(OH)2 --> Neutralization
Neutralization reactions produce a salt + H2O.
HCl + Mo(OH)2 ==> H2O + MoCl2
I know the formula for the salt to be MoCl2 because Cl must be -1 (since H is +1) and Mo must be +2 (since OH polyatomic ion is -1).
I look at the entire equation and see I have 2 Cl on the right so I can make that balance with a 2 for HCl.
2HCl + Mo(OH)2 ==> H2O + MoCl2
That balances the Cl. The Mo is ok. See if the 2HCl takes care of things. I have 4 H on the left (2 from 2HCl and 2 from Mo(OH)2 and that means we need a 2 for H2O on the right.
2HCl + Mo(OH)2 ==> 2H2O + MoCl2
Now we check it.
4 H on left and 4 on right.
2Cl on left and 2 on right.
1 Mo on left and 1 on right.
2 O on left and 2 on right.
zero charge on left and zero on right.