Question

Составить молекулярное и ионно-молекулярное уравнение! 1)AlCl3+Pb(NO3)2-> 2)Zn(OH)2+KOH->

Answer 1

2AlCl3 +3 Pb(NO3)2 = 2Al(NO3)3 + 3PbCl2

2Al(+3) + 6Cl(-) + 3Pb(+2) + 6NO3(-) = 2Al(+3) = 6NO3(-) + 3PbCl2

3Pb(+2) + 6Cl(-) = 3PbCl2

Zn(OH)2 +2 KOH =K2[Zn(OH)4]

Zn(OH)2 + 2K(+) + 2OH(-) = 2K(+) + [Zn(OH)4](-2)

Zn(OH)2 + 2OH(-) = [Zn(OH)4] (-2)

Answer 2

Molecular Equation:

The molecular equation for the given reactions can be written as follows:

1) AlCl3 + Pb(NO3)2 -> 2) Zn(OH)2 + KOH ->

Ionic Equation:

To write the ionic equation, we need to break down the reactants and products into their respective ions.

1) AlCl3 + Pb(NO3)2 -> - AlCl3 dissociates into Al3+ and Cl- ions. - Pb(NO3)2 dissociates into Pb2+ and NO3- ions.

The ionic equation can be written as: Al3+ + 3Cl- + Pb2+ + 2NO3- ->

2) Zn(OH)2 + KOH -> - Zn(OH)2 dissociates into Zn2+ and 2OH- ions. - KOH dissociates into K+ and OH- ions.

The ionic equation can be written as: Zn2+ + 2OH- + K+ + OH- ->

Please note that the equations provided above are not balanced. To obtain the balanced equations, you would need to balance the number of atoms on both sides of the equation.