• lead(II) nitrate = Pb(NO3)2 Sodium Sulfate = Na2(SO4) lead(II) sulfate = Pb(SO4) So we have: Pb(NO3)2 + Na2(SO4) --> Pb(SO4) + 2Na(NO3) Since we had two nitrates and two sodium on the right, we needed two of each on the left. Other wise this is balanced. Nitrate has a minus one charge so two of them are needed when put with a lead (II). Sulfate has a minus 2 charge so two sodium are needed. Sodium is only a plus 1. You may be required to include phases. Don't leave those out!

Copyright 2020, Wired Ivy, LLC

Answerbag | Terms of Service | Privacy Policy