The word equation would be iron reacts with lead nitrate to form iron two nitrate and lead.
Or perhaps you meant this?
The word equation would be Iron + Lead Nitrate => Iron (II) Nitrate + Lead
The chemical equation would be Fe(s) + Pb(NO3)2 => Fe(NO3)2 + Pb
I assume you are having problems with the charge the iron nitrate would go into, so here is a link that helps describe how to determine the charges of the ions of transitional metals: http://indiescience.org/basetalk/index.php?topic=13.0
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
3AgNO3 + FeCl3 ---> 3AgCl + Fe(NO3)3
Yes, of course: 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 for lead (IV) nitrate and 2 (NH4)3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)4 + 6 NH4NO3 for lead (II) nitrate.
If the copper nitrate formed is copper (I) nitrate, the equation balances with one atom of each metal and one formula weight of each nitrate. If the copper nitrate formed is copper (II) nitrate, the balanced equation is: 2 AgNO3 + Cu -> 2 Ag + Cu(NO3)2.
nigggers
Equation is Zn + Pb(NO3)2 --> Zn(NO3)2 + Pb
Zn2 + Fe3NO3 -> Zn3NO2 + Fe3
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Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
3AgNO3 + FeCl3 ---> 3AgCl + Fe(NO3)3
lead nitrate and hydrogen
Yes, of course: 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 for lead (IV) nitrate and 2 (NH4)3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)4 + 6 NH4NO3 for lead (II) nitrate.
Iron(III) Nitrate + Sodium Bicarbonate ----> Iron(III) Carbonate + Sodium Nitrate + Water + Carbon Dioxide2 Fe(NO3)3 + 6 NaHCO3 ----> Fe2(CO3)3 + 6 NaNO3 + 3 H2O + 3 CO2
Iron plus chlorine equals Iron chloride is the word equation.
If the copper nitrate formed is copper (I) nitrate, the equation balances with one atom of each metal and one formula weight of each nitrate. If the copper nitrate formed is copper (II) nitrate, the balanced equation is: 2 AgNO3 + Cu -> 2 Ag + Cu(NO3)2.
its already balanced
lead nitrate and water