Solving an Ionic Equation: Na2Co3 + H2S04 -> Na2S04 + H20 + C02

[dagg3r]

hi can somebody show me how to do this ionic equation thanks.


first of all the equation is

Na2Co3 + H2s04 -> Na2S04 + H20 + C02

i think the main parts are H20(l) and CO2(g) how do i do an ionic equation would i do

Co32- + 2H+ -> H20(l) + Co2(g)

please tell me if this is right thanks

 

[whozum]

You want to break down each component on either side to its cation and anion. This will help you identify spectator ions and the active ions in the process.

This is a simplier problem since each compound is formed of two major ions, for example the break down for $$Na_2CO_3$$ would be $$2Na^+ + CO_3^-$$

Go through the rest of the equation breaking the molecule up into its ions. Be careful on the products side, as some of them can be tricky.

 

[dagg3r]

check to see if active ions

ok i did that before i just want to know is Co2 gas and h20 water the active ions on the right hand side as Na2So4 is aqueous so it is a spectator?

 

[dextercioby]

Since $CO_{2}$ is a gas (so it doesn't stay in the aqueous solution),it think your reaction boils down to

$$CO_{3}^{2-}+2H^{+}\rightarrow CO_{2}\uparrow +H_{2}O$$

U see,i didn't write the spectator ions (the sulphate & the sodium ion).

Daniel.

 

[The Bob]

$$CO_{3}^{2-}+2H^{+}\rightarrow CO_{2}\uparrow +H_{2}O$$

And the arrow is standard practise to show that a gas is realized?

The Bob (2004 ©)

 

[ShawnD]

Generally you subscript the state of the ion/substance. CO3 is in solution so it's (aq), H+ is also in solution so it's (aq), CO2 is a gas so it's (g), water is a liquid so it's (l).

I've never had to write this before, but what if alcohol was mixed with water. Would the alcohol be (l) or (aq)?

 

[dextercioby]

Maybe the alcohol was in solid state before disolving it in water,so that "l" would not be accurate...

Daniel.

 

[ShawnD]

Alcohol freezes at a much lower temperature than water, so if you put solid alcohol in water you would get a weird solid alcohol >> liquid alcohol >> solid water ball that wouldn't exactly be "dissolved" in any sense. The same thing happens with dry ice in water.

 

[dextercioby]

That's not as interesting as the following case:pour pure (ethylic) alcohol at -20° in water.Should one get an aqueous solution of alcohol...?

Daniel.