Strontium nitrate and sulfuric acid equation

[sunbuster]

Find the molecular, total ionic, and net ionic equation for the aqueous reaction between strontium nitrate and sulfuric acid.

I did like this, please check this out...

Sr(NO3)2 + H2SO4 -> 2HNO3 + SrSO4 is molecular equation.

Sr+2(aq) + 2NO3-1(aq) + H+1(aq) + HSO4-1(aq) -> 2H+1(aq) + 2NO3-1(aq) + SrSO4(s) is total ionic equation.

Sr+2(aq) + HSO4-1(aq) -> H+1(aq) + SrSO4(s) is net ionic equation.

Are they correct? I'm really confusing the last one. The hydrogen can be canceled out or not?

Please help me out...! Thanks...

 

[QuarkCharmer]

You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.

 

[sunbuster]

You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.

=>>> I already know that part. I just want to check the final ionic equation. Anyway I edited.

 

[QuarkCharmer]

The 2 hydrogen ions and 2 nitrate ions on both side of the equation should cancel out. So you should have only the ions that took place in the reacion in the form of a synthesis.

Like: X+ + Y- -> XY

On the left side of your net equation you have Hydrogen sulfate (aq), since it is aqueous, it should be a hydrogen ion and a sulfate ion, then the hydrogen would cancel the one on the product side.

 

[sunbuster]

You should write the charges and states of each ion in the Net and Total ionic equations. You do this because you want to know what is aqueous and what is not. In the end equation there you should end up with one of those being a precipitate (solid) and then the finished equation would have only the two ions on the reactant side and the non aqueous product on the product side. If all of the molecules in the double replacement reaction are aqueous and no state change took place, then all of the ions would be considered spectator ions and thus the Net Ionic equation would actually be "No reaction" since all of the ions cancel out.

The 2 hydrogen ions and 2 nitrate ions on both side of the equation should cancel out. So you should have only the ions that took place in the reacion in the form of a synthesis.

Like: X+ + Y- -> XY

=>>> I think H2SO4(aq) is dissociate like H+(aq) + HSO4-(aq).
How 2 hydrogens should be canceled out? And I canceled out 2NO3 each side.

 

[QuarkCharmer]

Let's take a look at what exactly is happening here.

You have Sr(NO3)2 (aq) and you pour in H2(SO4)(aq). The double replacement rxn creates 2H(NO3)(aq) and Sr(SO4)(s).

The Net ionic equation (or end result) is that an Sr(aq) ion combines with an SO4(aq) ion to create Sr(SO4)(s).

Anything that is not reacting to produce the precipitate is considered a spectator and canceled out.

In this case, the 2 Hydrogen ions on the left (H2 is really 2 H atoms right?) are not taking place in the reaction at all, and therefor, they are present on both sides in aqueous form in the Total Ionic Equation. The nitrate (NO3) is also not taking place in the reaction to produce the precipitate, and as expected, in the Total Ionic Equation, there are the two nitrates on both sides. What you are left with is a solid of Strontium Sulfate and a bunch of Hydrogen and Nitrate ions floating about. But the purpose of the Net ionic is to show only those ions that were involved in the reaction to make the precipitate.

In your net ionic you have:

Sr+2(aq) + 2HSO4-1(aq) -> H+1(aq) + SrSO4(s)

The red is what is actually taking place. The polyatomic ion SO4 does not have a charge of -1, otherwise how could it initially be bonded to the diatomic molecule of Hydrogen (+2)?

 

[sunbuster]

Let's take a look at what exactly is happening here.

You have Sr(NO3)2 (aq) and you pour in H2(SO4)(aq). The double replacement rxn creates 2H(NO3)(aq) and Sr(SO4)(s).

The Net ionic equation (or end result) is that an Sr(aq) ion combines with an SO4(aq) ion to create Sr(SO4)(s).

Anything that is not reacting to produce the precipitate is considered a spectator and canceled out.

In this case, the 2 Hydrogen ions on the left (H2 is really 2 H atoms right?) are not taking place in the reaction at all, and therefor, they are present on both sides in aqueous form in the Total Ionic Equation. The nitrate (NO3) is also not taking place in the reaction to produce the precipitate, and as expected, in the Total Ionic Equation, there are the two nitrates on both sides. What you are left with is a solid of Strontium Sulfate and a bunch of Hydrogen and Nitrate ions floating about. But the purpose of the Net ionic is to show only those ions that were involved in the reaction to make the precipitate.

In your net ionic you have:

Sr+2(aq) + 2HSO4-1(aq) -> H+1(aq) + SrSO4(s)

The red is what is actually taking place. The polyatomic ion SO4 does not have a charge of -1, otherwise how could it initially be bonded to the diatomic molecule of Hydrogen (+2)?

=>>> Thanks a lot... you helped me so much... : ) but... I don't get the first problem about light. Do you remember the question about spectrum?

 

[QuarkCharmer]

I don't see a first problem about light?