Thursday
March 30, 2017

Post a New Question

Posted by on .

We did a lab where we reacted antacid with HCL and measured the pressure increase after the reaction. Then we found the percent concentrations of calcium carbonate of the antacid tablet. The question is: say the percent of calcium carbonate from trials 1 & 2 are consistent while trial 3 is much higher. How could a wet flask be used to explain this result? A wet flask would allow water to react with the calcium carbonate before we measure it, but i'm not sure what do say next. I think I need to use a gas law?

The setup: put a vial of HCl inside a flask of CaCO3. Plug the top with a stopper and suck out air to make it airtight, and there are instruments in the stopper that measure the inside temperature and pressure. When we are ready to measure, we tip the vial so that the HCl reacts with the CaCO3 to produce CO2 and since the flask should be airtight, the pressure will go up.

  • Chemistry Lab - ,

    Personally, I don't think a wet flask CAN account for high results (or low results, for that matter). The CaCO3 is essentially insoluble in water and won't react with any water there before the HCl vial is tipped. The CaCO3 can't tell the difference between water that might be there before the HCl vial is tipped and the water that's in the HCl solution. I think you need to look for other reasons that would account for the high results. Perhaps the tablet's mass was recorded or weighed incorrectly. Perhaps the flask was not cleaned properly between trials 2 and 3.
    Perhaps another tutor will give his/her opinion.

  • Chemistry Lab - ,

    Nevermind I think i figured it out. But CaCO3 is supposed to react with water in the presence of CO2

  • Chemistry Lab - ,

    CO2(g) + CaCO3(s)==> Ca(HCO3)2(aq)

  • Chemistry Lab - ,

    action with acid

  • Chemistry Lab - ,

    properties of hydrogen

  • Chemistry Lab - ,

    can you please send your answer here ?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question