# Ionic equations help.

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Well basically I've started AS and I'm struggling quite a lot. But anyway we got a questions asking us to 'write out an ionic equation for the reaction of magnesium with sulfuric acid'

So I got Mg(2+) + H2(2+)SO4(2-) ----> Mg2SO4 + H2.

Please help me.

Thank you.

I'm a tad confused and really worried because if I struggle with this, how am I going to fare with other topics?

So I got Mg(2+) + H2(2+)SO4(2-) ----> Mg2SO4 + H2.

Please help me.

Thank you.

I'm a tad confused and really worried because if I struggle with this, how am I going to fare with other topics?

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#3

(Original post by

Well basically I've started AS and I'm struggling quite a lot. But anyway we got a questions asking us to 'write out an ionic equation for the reaction of magnesium with sulfuric acid'

So I got Mg(2+) + H2(2+)SO4(2-) ----> Mg2SO4 + H2.

Please help me.

Thank you.

I'm a tad confused and really worried because if I struggle with this, how am I going to fare with other topics?

**senz72**)Well basically I've started AS and I'm struggling quite a lot. But anyway we got a questions asking us to 'write out an ionic equation for the reaction of magnesium with sulfuric acid'

So I got Mg(2+) + H2(2+)SO4(2-) ----> Mg2SO4 + H2.

Please help me.

Thank you.

I'm a tad confused and really worried because if I struggle with this, how am I going to fare with other topics?

First of all in an equation make sure the state symbols are in. Then if the same compound, e.g. so4 comes up in both the reactants and the product, then it is a spectator and is NOT meant to be included in your ionic equation.

Give me the original equation and the state symbols then i'll help you

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(Original post by

First of all in an equation make sure the state symbols are in. Then if the same compound, e.g. so4 comes up in both the reactants and the product, then it is a spectator and is NOT meant to be included in your ionic equation.

Give me the original equation and the state symbols then i'll help you

**EffKayy**)First of all in an equation make sure the state symbols are in. Then if the same compound, e.g. so4 comes up in both the reactants and the product, then it is a spectator and is NOT meant to be included in your ionic equation.

Give me the original equation and the state symbols then i'll help you

Write out an ionic equation for the reaction of magnesium with sulfuric acid.

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#5

You're worrying too much and you have to wait more than a few minutes for a reply normally

Step 1, write the full equation: Mg

step 2, split up the aqueous components into ions: Mg

step 3: cancel ions that appear on both sides, can you finish that bit?

Step 1, write the full equation: Mg

_{(s)}+ H_{2}SO_{4(aq)}---> MgSO_{4(aq)}+ H_{2(g)}step 2, split up the aqueous components into ions: Mg

_{(s)}+ 2H^{+}_{(aq)}+ SO_{4}^{2-}_{(aq)}---> Mg^{2+}_{(aq)}+ SO_{4}^{2-}_{(aq)}+ H_{2(g)}step 3: cancel ions that appear on both sides, can you finish that bit?

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(Original post by

You're worrying too much and you have to wait more than a few minutes for a reply normally

Step 1, write the full equation: Mg

step 2, split up the aqueous components into ions: Mg

step 3: cancel ions that appear on both sides, can you finish that bit?

**EierVonSatan**)You're worrying too much and you have to wait more than a few minutes for a reply normally

Step 1, write the full equation: Mg

_{(s)}+ H_{2}SO_{4(aq)}---> MgSO_{4(aq)}+ H_{2(g)}step 2, split up the aqueous components into ions: Mg

_{(s)}+ 2H^{+}_{(aq)}+ SO^{2-}_{4(aq)}---> Mg^{2+}_{(aq)}+ SO^{2-}_{4(aq)}+ H_{2(g)}step 3: cancel ions that appear on both sides, can you finish that bit?

Sorry about this.

P.S It asks for the full, balanced ionic equation.

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#7

(Original post by

So Mg + 2H+ --> Mg+ H2?

Sorry about this.

P.S It asks for the full, balanced ionic equation.

**senz72**)So Mg + 2H+ --> Mg+ H2?

Sorry about this.

P.S It asks for the full, balanced ionic equation.

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(Original post by

You just need the state symbols and the charge on Mg on the right

**EierVonSatan**)You just need the state symbols and the charge on Mg on the right

Could you help me with one or two more problems?

Feel free to say no.

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#9

(Original post by

Thank you.

Could you help me with one or two more problems?

Feel free to say no.

**senz72**)Thank you.

Could you help me with one or two more problems?

Feel free to say no.

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#10

(Original post by

Just post your questions and someone will answer them, it's what the forum is for

**EierVonSatan**)Just post your questions and someone will answer them, it's what the forum is for

I'm trying to work through some of her problems and I understand the idea of...erm..."cancelling out" terms like you can in algebra, but I haven't managed to get correct answers yet.

- Do you only split the aqueous components of the equation into individual terms (with + signs between, I mean)?
- How is it possible to identify which are aqueous? Is it on the basis of there being in there?
- How do charges work? Is 2Na the same as (in that both have a +2 effect)? If so, would you write "2+" for the "power" in both cases? (Sorry for the Maths analogies - it's the only thing like this I'm really familiar with )
- If there are several of a component on one side of the equation and only one on the other, can you "subtract" that one component or not?

To give you an example of the kind of thing I'm working with:

I've broken down the aqueous components (but I'll omit the charge notation since I'm not sure how it works):

There's on the left and on the right - can I subtract from both sides? (I don't think so but decided I'd better check...)

Thanks so much for your help - this is due tomorrow () so I'd appreciate any insight!

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#11

(Original post by

Hi, I've got a few questions about ionic equations too (hopefully OP doesn't mind me using this thread rather than starting a new one!). I'm not a chemist but my sister's doing AS and she doesn't understand this either.

I'm trying to work through some of her problems and I understand the idea of...erm..."cancelling out" terms like you can in algebra, but I haven't managed to get correct answers yet.

**Tortious**)Hi, I've got a few questions about ionic equations too (hopefully OP doesn't mind me using this thread rather than starting a new one!). I'm not a chemist but my sister's doing AS and she doesn't understand this either.

I'm trying to work through some of her problems and I understand the idea of...erm..."cancelling out" terms like you can in algebra, but I haven't managed to get correct answers yet.

Answers in bold:

- Do you only split the aqueous components of the equation into individual terms (with + signs between, I mean)?
**Yep** - How is it possible to identify which are aqueous? Is it on the basis of there being in there?
**Sometimes it's given, other times it'll be implied** - How do charges work? Is 2Na the same as (in that both have a +2 effect)? If so, would you write "2+" for the "power" in both cases? (Sorry for the Maths analogies - it's the only thing like this I'm really familiar with )
**Nope, 2Na means (2 x Na) while Na**_{2}is more like ''Na^{2}'' if the algebra analogy helps i.e. 2Na is two sodium atoms while Na_{2}is two sodium atoms attached to each other in a molecule... - If there are several of a component on one side of the equation and only one on the other, can you "subtract" that one component or not?
**Yes such as 2A + B ----> C + A becomes A + B ---> C**

2H

^{+}+ SO

_{4}

^{2-}+ 2Na

^{+}+ 2OH

^{-}---> 2H

_{2}O + 2Na

^{+}+ SO

_{4}

^{2-}

Importantly water is not an ionic compound and should not be split up into ions...

Can you now see the equivalent ''terms'' on both sides?

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#12

(Original post by

The charges are all important as it describes the way the molecules are built it should be:

2H

Importantly water is not an ionic compound and should not be split up into ions...

**EierVonSatan**)The charges are all important as it describes the way the molecules are built it should be:

2H

^{+}+ SO_{4}^{2-}+ 2Na^{+}+ OH^{-}---> 2H_{2}O + 2Na^{+}+ SO_{4}^{2-}Importantly water is not an ionic compound and should not be split up into ions...

Are the charges supposed to cancel out (on each side of the equation)? From the little Chem I do remember, I know H and Na are in Group 1, so I'd have thought that 2H

^{+}represents 2 "lots of" positive charge. As such, adding the charges, I'd get (+2) + (-2) + (+2) + (-1) = +1. Or do you not "multiply" the charge by the coefficient to take into account the magnitude?

Thanks again.

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#13

(Original post by

Cheers - answering my initial questions has certainly helped to make things clearer. However, I'm not following your last few steps.

Are the charges supposed to cancel out (on each side of the equation)? From the little Chem I do remember, I know H and Na are in Group 1, so I'd have thought that 2H

Thanks again.

**Tortious**)Cheers - answering my initial questions has certainly helped to make things clearer. However, I'm not following your last few steps.

Are the charges supposed to cancel out (on each side of the equation)? From the little Chem I do remember, I know H and Na are in Group 1, so I'd have thought that 2H

^{+}represents 2 "lots of" positive charge. As such, adding the charges, I'd get (+2) + (-2) + (+2) + (-1) = +1. Or do you not "multiply" the charge by the coefficient to take into account the magnitude?Thanks again.

On both sides of a chemical equation two things need to be equal: the mass and the charge - thus you can't have more charge on one side than the other.

There are two main compound types - covalent species and ionic ones, the former will not split up into ions when dissolved (water is such an example). The latter will as they're made up from ions (atoms with charges) such as NaCl.

Undissolved NaCl is an ionic solid, add some water and it splits up into it's two sets of ions Na

^{+}and Cl

^{-}, these ions are formed because they lose/gain electrons to fill their outermost shell (i.e. to be more stable). This can be written: NaCl ---> Na

^{+}+ Cl

^{-}(mass and charge balanced).

2H

^{+}means 2 x H

^{+}= total charge of +2

for the example: 2H

^{+}

**(+2)**+ SO

_{4}

^{2-}

**(-2)**+ 2Na

^{+}

**(+2)**+ 2OH

^{-}

**(-2)**---> 2H

_{2}O + 2Na

^{+}

**(+2)**+ SO

_{4}

^{2-}

**(-2)**

0 = 0 i.e. overall neutral, as we'd expect a full equation to be

p.s. sorry, I missed the 2 in front of the OH

^{-}above

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#15

(Original post by

I need to know the ionic equation for sulfuric acid... anyone help?

**Chloeisabelle_**)I need to know the ionic equation for sulfuric acid... anyone help?

Sulfuric acid doesn't have an "ionic equation": it has a formula. It is made of ions and (some of) these can be shown in an ionic equation, when it reacts with something else.

What are you reacting it with?

Last edited by Pigster; 4 weeks ago

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#17

**senz72**)

Well basically I've started AS and I'm struggling quite a lot. But anyway we got a questions asking us to 'write out an ionic equation for the reaction of magnesium with sulfuric acid'

So I got Mg(2+) + H2(2+)SO4(2-) ----> Mg2SO4 + H2.

Please help me.

Thank you.

I'm a tad confused and really worried because if I struggle with this, how am I going to fare with other topics?

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#18

(Original post by

you probably won't see this but i have started year 12 this september and i'm pretty i've got given the exact same sheet as you did 9 years ago - i just think it's funny how i literally have the same problem as you did because i don't get it all LMAO

**ishie889**)you probably won't see this but i have started year 12 this september and i'm pretty i've got given the exact same sheet as you did 9 years ago - i just think it's funny how i literally have the same problem as you did because i don't get it all LMAO

I'll bet that it won't be the same worksheet. That equation is a fairly standard example as an introduction to ionic equations.

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