Aluminium forms two oxdiation states with the chloride ion, I and III
3AlCl + Na3PO4 ------------> NaCl + Al3PO4 this is for Aluminum I chloride
AlCl3 + Na3PO4 ------------> AlPO4 + 3 NaCL for the Aluminum III chloride
This purely is elementary Chemistry, only a Chemistry noob would not know how to do this.
Noobness aside, the answer is 3Al + 3NH4ClO4 --> Al2O3 + AlCl3 + 3NO + 6H2O
alcl4+cl3
what are the products of the reaction between aluminum chloride and cesium
whats the balanced chemical equation
The balanced equation is: Zn(C2H3O2)2 + NaPO4 --> ZnPO4 + Na(C2H3O2)2.
4 LiH + AlCl3 =======> LiAlH4 + 3 LiCl
2Na3(PO4) + 3Pb(II)Cl2 --> 6NaCl + Pb3(PO4)2
what are the products of the reaction between aluminum chloride and cesium
aluminium chloride and oxygen gas
whats the balanced chemical equation
3Ca + 2AlCl3 ----> 3CaCl2 + 2Al
The balanced equation is: Zn(C2H3O2)2 + NaPO4 --> ZnPO4 + Na(C2H3O2)2.
2 AlCl3 -> 2 Al + 3 Cl2
4 LiH + AlCl3 =======> LiAlH4 + 3 LiCl
2Na3(PO4) + 3Pb(II)Cl2 --> 6NaCl + Pb3(PO4)2
3Ni(NO2)2 (aq) + 2Na3PO4 (aq) = Ni3(PO4)2 (s) + 6NaNO3 (aq)
Al(3+) + Cl(1-) = AlCl3
oxygen
2Na3PO4 + 3CaCl2 --> Ca3(PO4)2 + 6NaCl