Find the amount of moles of carbon dioxide produced during the reaction

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SUMMARY

The discussion focuses on calculating the amount of moles of carbon dioxide (CO2) produced from the reaction of calcium carbonate (CaCO3) with hydrochloric acid (HCl). The molar mass of CaCO3 is determined to be 100 g/mol, while the molar mass of CO2 is 44 g/mol. Using the formula for moles (n = m/M_r), it is established that 5 grams of CaCO3 yields 0.05 moles of CO2, based on the balanced reaction equation: CaCO3 + 2HCl → CaCl2 + CO2 + H2O. Therefore, 0.05 moles of CO2 are produced in this reaction.

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  • Knowledge of chemical reaction balancing
  • Basic concepts of moles in chemistry
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  • Study the concept of molar mass and its significance in chemical reactions
  • Learn about stoichiometric coefficients in balanced chemical equations
  • Explore the properties and reactions of carbonates with acids
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Chemistry students, educators, and anyone involved in chemical calculations or reactions involving carbonates and acids will benefit from this discussion.

mathlearn
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i.If all the calcium carbonate used in set up X was used up for the reaction, what is the amount of moles of carbon dioxide produced during the reaction ? (Ca = 40, C = 12, O = 16)

My progress:

Molar mass of CaCO3=(40+12+(16*3))gmol-1=100 gmol-1

Molar mass of CO2=12 + 16* 2 gmol-1=44 gmol-1

After that what must be done ? :confused:

Many Thanks :)
 

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mathlearn said:
i.If all the calcium carbonate used in set up X was used up for the reaction, what is the amount of moles of carbon dioxide produced during the reaction ? (Ca = 40, C = 12, O = 16)

My progress:

Molar mass of CaCO3=(40+12+(16*3))gmol-1=100 gmol-1

Molar mass of CO2=12 + 16* 2 gmol-1=44 gmol-1

After that what must be done ? :confused:

Many Thanks :)

Now you've worked out the molar mass of your substances you can work out the amount of moles of CaCO3 using the equation [math]n = \dfrac{m}{M_r} \text{ or } \text{moles} = \dfrac{\text{mass}}{\text{molar mass}}[/math]

[math]n_{CaCO_3} = \dfrac{5}{100} = \dfrac{1}{20} = 0.05 \text{mol}[/math]Next consider the balanced reaction that is taking place here which is that of a carbonate with acid

[math]CaCO_{3 (s)} + 2HCl_{(aq)} \rightarrow CaCl_{2 (s)} + CO_{2 (g)} + H_2O_{(l)}[/math]

From the equation above we see that one mole of [math]CaCO_3[/math] produces one mole of [math]CO_2[/math] which means you'll have [math]0.05[/math] moles of [math]CO_2[/math] which is the answer to your question.
 

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