Ionic equation for barium chloride and sulphuric acid? Watch

cardboardoranges
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#1
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is it bacl2 + so4 -> baso4 + hcl
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Tarick
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I believe it is
BaCl2 + H2SO4 2- --> BaSO4 + 2HCl
Barium Chloride Charge is Nil
Sulphuric Acid Charge Is 2-
Hydrochloric Acid is a molecule so it cannot form ionic bonds.
I dont know if im right im in Y9
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Dylann
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(Original post by cardboardoranges)
is it bacl2 + so4 -> baso4 + hcl

(Original post by Tarick)
I believe it is
BaCl2 + H2SO4 2- --> BaSO4 + 2HCl
Barium Chloride Charge is Nil
Sulphuric Acid Charge Is 2-
Hydrochloric Acid is a molecule so it cannot form ionic bonds.
I dont know if im right im in Y9
The sulphate ion (SO4) has the 2- charge, not the molecule. All molecules are neutral.

As for the question, it's an ionic equation not a normal balanced equation, so we have to do things a little differently...

First write out all the reactants and products in their states:

Ba2+ (aq) + 2Cl- (aq) + 2H+ (aq) + SO4 2- (aq) --> Ba2+ (s) + SO4 2- (s) + 2H+ (aq) + 2Cl- (aq)

You need to know that BaSO4 is insoluble so the state changes from (aq) to (s).

Now you cancel out the ions that are exactly the same on each side (exactly the same means same state too).

Ba2+ (aq) + 2Cl- (aq) + 2H+ (aq) + SO4 2- (aq) --> Ba2+ (s) + SO4 2- (s) + 2H+ (aq) + 2Cl- (aq)

To finish off, write BaSO4 as a molecule because it doesn't dissociate into separate ions (since it's a solid):

Ba2+ (aq) + SO4 2- (aq) ---> BaSO4 (s)

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cardboardoranges
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(Original post by Dylann)
The sulphate ion (SO4) has the 2- charge, not the molecule. All molecules are neutral.

As for the question, it's an ionic equation not a normal balanced equation, so we have to do things a little differently...

First write out all the reactants and products in their states:

Ba2+ (aq) + 2Cl- (aq) + 2H+ (aq) + SO4 2- (aq) --> Ba2+ (s) + SO4 2- (s) + 2H+ (aq) + 2Cl- (aq)

You need to know that BaSO4 is insoluble so the state changes from (aq) to (s).

Now you cancel out the ions that are exactly the same on each side (exactly the same means same state too).

Ba2+ (aq) + 2Cl- (aq) + 2H+ (aq) + SO4 2- (aq) --> Ba2+ (s) + SO4 2- (s) + 2H+ (aq) + 2Cl- (aq)

To finish off, write BaSO4 as a molecule because it doesn't dissociate into separate ions (since it's a solid):

Ba2+ (aq) + SO4 2- (aq) ---> BaSO4 (s)

Thabks so much.
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Mohkaam
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(Original post by dylann)
the sulphate ion (so4) has the 2- charge, not the molecule. All molecules are neutral.

As for the question, it's an ionic equation not a normal balanced equation, so we have to do things a little differently...

First write out all the reactants and products in their states:

Ba2+ (aq) + 2cl- (aq) + 2h+ (aq) + so4 2- (aq) --> ba2+ (s) + so4 2- (s) + 2h+ (aq) + 2cl- (aq)

you need to know that baso4 is insoluble so the state changes from (aq) to (s).

Now you cancel out the ions that are exactly the same on each side (exactly the same means same state too).

Ba2+ (aq) + 2cl- (aq) + 2h+ (aq) + so4 2- (aq) --> ba2+ (s) + so4 2- (s) + 2h+ (aq) + 2cl- (aq)

to finish off, write baso4 as a molecule because it doesn't dissociate into separate ions (since it's a solid):

Ba2+ (aq) + so4 2- (aq) ---> baso4 (s)

:d
perfect answer thank you
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