Question

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.

Answer 1

@toxicsugar22 @aaronq

Answer 2

@Abhisar @iambatman can you help?

Answer 3

Know the chemical equation to this reaction?

Answer 4

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

Answer 5

@Empty what do you do with the resulting weight....

Answer 6

Weight of reactant-resulting weight=weight of CO2 in g
Density is mass/volume.
They give you the density of CO2

Answer 7

Since carbon dioxide is released then measured

Answer 8

"Weight of reactants"
Forgot the s in reactants in my earlier post.

Answer 9

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

Answer 10

Yes, I got that.

Answer 11

\[\sf 0.23 ~g~ CO_2 ~\times \frac{1~L}{1.81~g~CO_2} = 0.127~L ~CO_2 =\boxed{ 0.13~L~CO_2} \]

Answer 12

Yea, that is right, but it should be in 3 sf since density of CO2 has the least (3sf)
Nice job!

Answer 13

Oh, I thought it was based on the lowest # of sf

Answer 14

Which is 0.127

Answer 15

Ahh, so it must match up to the density of carbon dioxide given.

Answer 16

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

Answer 17

Yeah, I got that.

Answer 18

Thanks for the help @jhannybean and @shalante...but how do you know that the resulting weight is all CO2???

Answer 19

\(\sf CO_2\) is a gas and is the only thing lost after the reaction happens

Answer 20

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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?