A 1.009g mixture of Na2SO4 (molar mass=142.04 g/mol) and Pb(NO3)2 (molar mass= 331.20 g/mol)?
A)write th3 molecular form of the equation for the reaction.
b)write the net ionic equation for the reaction.
Na2SO4(aq) + Pb(NO3)2(aq) ==> PbSO4(s) + 2NaNO3(aq)
SO4^2-(aq) + Pb^2+(aq) ==> PbSO4(s)
To determine the molecular and net ionic equations for the reaction between Na2SO4 and Pb(NO3)2, we need to first understand the reaction that occurs. The reactants are Na2SO4 and Pb(NO3)2, and the products are formed by swapping the cations from each compound to create new compounds.
a) Writing the molecular equation:
The molecular equation shows the chemical reaction as it is written without any consideration of the individual ions. We can use the following notation to represent the reactants and products:
Na2SO4 + Pb(NO3)2 -> NaNO3 + PbSO4
In the reaction, Na2SO4 swaps its sodium ion (Na+) with the lead ion (Pb2+) in Pb(NO3)2 to form sodium nitrate (NaNO3) and lead sulfate (PbSO4).
b) Writing the net ionic equation:
The net ionic equation shows only the ions involved in the reaction, excluding any spectator ions that do not take part in the chemical change.
To determine the net ionic equation, we need to dissociate all the compounds into their respective ions. Sodium sulfate (Na2SO4) dissociates into two sodium ions (2Na+) and one sulfate ion (SO42-). Lead nitrate (Pb(NO3)2) dissociates into one lead ion (Pb2+) and two nitrate ions (2NO3-).
So, the ionic form of the reaction is:
2Na+(aq) + SO42-(aq) + Pb2+(aq) + 2NO3-(aq) -> 2Na+(aq) + 2NO3-(aq) + PbSO4(s)
Now, we can cancel out the spectator ions (2Na+ and 2NO3-) that appear on both sides of the equation:
SO42-(aq) + Pb2+(aq) -> PbSO4(s)
This is the net ionic equation for the reaction, which shows that the sulfate ion (SO42-) combines with the lead ion (Pb2+) to form solid lead sulfate (PbSO4).
Please note that the states of matter for each compound or ion (e.g., (aq) for aqueous, (s) for solid) are usually included in such equations but were omitted in this case.