Post a New Question

Chemsitry

posted by .

Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of grams of CaCO3 needed to exactly react with 250 mL of stomach acid.
CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ H20(l)

  • Chemsitry -

    I answered this below.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    How many milliliters of 0.418 M HCl are needed to react with 52.7 g of CaCO3?
  2. chemistry

    Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric acid) according to the following equation: CaCO3(s)+2HCl(aq)->CaCl2(aq)+H2O(l)+CO2(g) Tums, an antacid, contains CaCO3. If Tums is added to 35.0 mL of 0.300 M HCl,how …
  3. Chemistry

    Calcium carbonate reacts w/stomach acid according to the following chemical equation. CaCO3+2HCl(aq)-> CaCl2(aq)+H2O(l)+CO2(g) A. Balance the eqaution B. Tums is one commercially sold antacid that contains CACO3. If Tums is added …
  4. chemistry

    How many milliliters of 0.237 M HCl are needed to react with 52.2g of CaCO3?
  5. Chemistry

    How many grams of calcium carbonate, CaCO3, do you need in an antacid tablet to neutralize 20 mL of stomach acid?
  6. Chemistry

    Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of CaCO3 needed to exactly react with 250 mL of stomach acid. CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ …
  7. Chemistry

    4. A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl. It took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid. a) Calculate the volume of HCl neutralized by the NaOH. b) Calculate …
  8. Chemistry

    If a tablet of TUMS contains 500mg of the active ingredient, CaCO3, how many tablets would it take to neturalize the 1.5g of HCl?
  9. Chemistry

    Tums tablets contain CaCO3 that reacts with stomach acid as follows: CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)  Suppose the tablet was analyzed by adding 50.00 mL of 0.200 M HCl, which resulted in leftover …
  10. chemistry

    A 5.309 g antacid tablet, with CaCO3 as the active ingredient, was mixed was 30.0 mL of 0.831 M HCl. After the reaction occurred, it took 15.3 mL of 0.7034 M NaOH to neutralize the excess acid. a) How much HCl (in mL) of was neutralized …

More Similar Questions

Post a New Question