[omo5031]

25.0cm^3 of a solution containing 80.0g of sodium hydrogen-carbonate per dm^3 of solution was titrated against 0.50 molar sulfuric acid.
1) write down the equation for this reaction.
2) what is the molarity of the sodium hydrogen-carbonate solution?
3) what volume of carbon dioxide, measured at stp would be evolved during the reaction?
4) calculate the volume of sulfuric acid required to neutralize the sodium hydrogen-carbonate solution.

can anyone please show me how they got the answers. Thanks in advance

[Arkcon]

See link above: Forum Rules: Read This Before Posting
Well, lets get started.  Part one asks for a balanced equation.  Can you begin to write that out?

[omo5031]

See link above: Forum Rules: Read This Before Posting
Well, lets get started.  Part one asks for a balanced equation.  Can you begin to write that out?

Yes I can, I need help with the other parts, here is the equation
   
2 NaHCO3(aq) + H2SO4(aq) --> Na2SO4(aq) + 2 CO2(g) + 2 H2O(aq)
 Please help with the rest

[Arkcon]

You should also balance the chemical equation that you wrote, so that it is useful to you later on.  opps, I double checked, that was balanced, srry

Now, for part two:  You been given a volume of liquid, and grams of solid.  You need molarity.  Can you calculate that?

[omo5031]

You should also balance the chemical equation that you wrote, so that it is useful to you later on.  opps, I double checked, that was balanced, srry

Now, for part two:  You been given a volume of liquid, and grams of solid.  You need molarity.  Can you calculate that?

No I did it but I'm not really sure about my answer. I got 0.952M. I divided the given mass by the RFM. But I'm not sure if this is right

[Arkcon]

OK.  I got that number as well.  I have no idea what RFM^* stands for, I divided by molecular mass of the solid.  Do you want to know why?  Look at this:

80.0 g of NaHCO3 X 1mole of NaHCO3/84.01 g NaHCO3 = 0.952 mole NaHCO3

By writing it out like this, I'm able to multiply by the grams given by what is essentially 1, the ration of grams to moles for this particular substance, cancelling out units, which I struck through.  This is one way to help keep track.  Note:  I've calculated moles, and the question asks for molarity.  If the answer correct, for this problem?  And why?

[omo5031]

OK.  I got that number as well.  I have no idea what RFM^* stands for, I divided by molecular mass of the solid.  Do you want to know why?  Look at this:

80.0 g of NaHCO3 X 1mole of NaHCO3/84.01 g NaHCO3 = 0.952 mole NaHCO3

By writing it out like this, I'm able to multiply by the grams given by what is essentially 1, the ration of grams to moles for this particular substance, cancelling out units, which I struck through.  This is one way to help keep track.  Note:  I've calculated moles, and the question asks for molarity.  If the answer correct, for this problem?  And why?

I think its correct. I'm not sure thats why I'm asking for help. Can you please do number 3, I have no idea at all on how to do that.

[sjb]

OK.  I got that number as well.  I have no idea what RFM^* stands for, I divided by molecular mass of the solid.  Do you want to know why?  Look at this:

80.0 g of NaHCO3 X 1mole of NaHCO3/84.01 g NaHCO3 = 0.952 mole NaHCO3

By writing it out like this, I'm able to multiply by the grams given by what is essentially 1, the ration of grams to moles for this particular substance, cancelling out units, which I struck through.  This is one way to help keep track.  Note:  I've calculated moles, and the question asks for molarity.  If the answer correct, for this problem?  And why?

I think its correct. I'm not sure thats why I'm asking for help. Can you please do number 3, I have no idea at all on how to do that.

If there were 0.952 mole of NaHCO₃ in the reaction, how many moles of CO₂ were given off? Do you know any formulae that relate volume to number of moles for a gas?

[omo5031]

OK.  I got that number as well.  I have no idea what RFM^* stands for, I divided by molecular mass of the solid.  Do you want to know why?  Look at this:

80.0 g of NaHCO3 X 1mole of NaHCO3/84.01 g NaHCO3 = 0.952 mole NaHCO3

By writing it out like this, I'm able to multiply by the grams given by what is essentially 1, the ration of grams to moles for this particular substance, cancelling out units, which I struck through.  This is one way to help keep track.  Note:  I've calculated moles, and the question asks for molarity.  If the answer correct, for this problem?  And why?

I think its correct. I'm not sure thats why I'm asking for help. Can you please do number 3, I have no idea at all on how to do that.

If there were 0.952 mole of NaHCO₃ in the reaction, how many moles of CO₂ were given off? Do you know any formulae that relate volume to number of moles for a gas?

No I don't can you please show me.

[vmelkon]

You wrote 2 NaHCO3(aq) + H2SO4(aq) --> Na2SO4(aq) + 2 CO2(g) + 2 H2O(aq)

so 2 moles of NaHCO3 give you 2 moles of CO2.
Therefore
0.952 moles of nahco3 * 2 moles of co2/2moles of nahco3
and the "moles of nahco3" cancel out. The answer is in "moles of co2".
ALWAYS write it in that format to make it clear.

Then you have to look up what STP means.
To know the volume of CO2, you have a choice:
1. use the ideal gas equation
2. use the real gas equation (called the van der Waals equation). a = 3.59 and b=0.0427