raffle_snaffle
  • raffle_snaffle
You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 64.96 g . The relevant equation is CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq) Assuming no other reactions take place, what mass of CO2 was produced in this reaction?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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raffle_snaffle
  • raffle_snaffle
@Photon336
Photon336
  • Photon336
\[CaCO_{3} = 100.1 gramg/mol\] \[HCl molar mass = 36.46 g/mol \] \[14.0 g CaCO_{3}*\frac{ mol }{ 100.1grams } = 0.14 moles CaCO_{3}\] \[56.70g HCl \frac{ mol }{ 36.46 } = 1.6 mol HCL\]
Photon336
  • Photon336
CaCO3 is the limiting reagent so we use that \[0.14 mol CaCO_{3} *((CO_{2})/(CaCO_{3}) = 0.14 mol CO_{3} * (44g/mol) = 6.2grams\]

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raffle_snaffle
  • raffle_snaffle
that is wrong. I solved it already. its 5.74g
raffle_snaffle
  • raffle_snaffle
suppose to use the law of conservation
raffle_snaffle
  • raffle_snaffle
14.00g + 56.70g = 70.70g total
raffle_snaffle
  • raffle_snaffle
70.70g - 64.96g = 5.74g lost of CO2
Photon336
  • Photon336
ah .. wait YES. i see.
Photon336
  • Photon336
CO2 bubbled out of your reaction so we did not need to even consider doing stioch to begin with. because the question states that there aren't any side reactions going on.
Photon336
  • Photon336
@raffle_snaffle I see. I had to look at this a second time. In our reaction, CO2 was being released so, that's where the loss of mass was from. technically if we had 70 grams starting we should have the same amount in our products.

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