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anonymous

  • one year ago

Tums tablets (contain Calcium Carbonate and other inert substances) when ingested react with gastric juice (HCl) in the stomach to give off CO2(g). When a 1.328 g tablet reacted with 40.00mL HCL (density 1.140g/mL), CO2 was given off and the resulting solution weighed 46.699g. Calculate the number of liters of CO2 was released if it's density is 1.81 g/L.

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  1. anonymous
    • one year ago
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    @toxicsugar22 @aaronq

  2. anonymous
    • one year ago
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    @Abhisar @iambatman can you help?

  3. anonymous
    • one year ago
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    Know the chemical equation to this reaction?

  4. Jhannybean
    • one year ago
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    \[\sf CaCO_3 ~(s) + 2HCl~ (aq) \longrightarrow CO_2 ~(g) +CaCl_2 ~(aq)+H_2O~(l)\] given: \[\sf reacted:1.328 \ \ g \ CaCO_3\]\[\sf reacted: 40.00 \ mL \ HCl \times \frac{1.140 ~ g~ HCl}{1 ~ mL~ HCl} = 45.60~ g~ HCl\]\[\sf solution ~ weight ~=46.699~ g\]

  5. Jhannybean
    • one year ago
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    @Empty what do you do with the resulting weight....

  6. anonymous
    • one year ago
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    Weight of reactant-resulting weight=weight of CO2 in g Density is mass/volume. They give you the density of CO2

  7. anonymous
    • one year ago
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    Since carbon dioxide is released then measured

  8. anonymous
    • one year ago
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    "Weight of reactants" Forgot the s in reactants in my earlier post.

  9. Jhannybean
    • one year ago
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    Okay, well... \[\sf weight~ of ~reactant - resulting ~weight \]\[\sf = (1.328 ~g ~CaCO_3~+~45.60~ g~ HCl) -~ 46.699~g\]\[\sf =46.928~g - 46.699~g \]\[\sf = 0.23~ g~ CO_2\]

  10. anonymous
    • one year ago
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    Yes, I got that.

  11. Jhannybean
    • one year ago
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    \[\sf 0.23 ~g~ CO_2 ~\times \frac{1~L}{1.81~g~CO_2} = 0.127~L ~CO_2 =\boxed{ 0.13~L~CO_2} \]

  12. anonymous
    • one year ago
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    Yea, that is right, but it should be in 3 sf since density of CO2 has the least (3sf) Nice job!

  13. Jhannybean
    • one year ago
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    Oh, I thought it was based on the lowest # of sf

  14. anonymous
    • one year ago
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    Which is 0.127

  15. Jhannybean
    • one year ago
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    Ahh, so it must match up to the density of carbon dioxide given.

  16. anonymous
    • one year ago
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    0.13 is 2 sf since, since 0 does not count as a sf 0.127 has 3 sf since 0 does not count as a sf

  17. Jhannybean
    • one year ago
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    Yeah, I got that.

  18. anonymous
    • one year ago
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    Thanks for the help @jhannybean and @shalante...but how do you know that the resulting weight is all CO2???

  19. Jhannybean
    • one year ago
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    \(\sf CO_2\) is a gas and is the only thing lost after the reaction happens

  20. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean Okay, well... \[\sf weight~ of ~reactant - resulting ~weight\]\[\sf = (1.328 ~g ~CaCO_3~+~45.60~ g~ HCl) -~ 46.699~g\]\[\sf =46.928~g - 46.699~g\]\[\sf = 0.23~ g~ CO_2\] \(\color{#0cbb34}{\text{End of Quote}}\) This is the initial weight of both reactants when they react minus the weight of the resulting solution (minus the \(\sf CO_2\)); so the weight you get is of the \(\rm CO_2\) lost. Does that make sense?

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