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

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The discussion focuses on calculating the moles of carbon dioxide produced from the reaction of calcium carbonate (CaCO3) with hydrochloric acid (HCl). The molar mass of CaCO3 is determined to be 100 g/mol, and the molar mass of CO2 is 44 g/mol. Using the formula for moles, it is established that 5 grams of CaCO3 corresponds to 0.05 moles. The balanced chemical equation indicates that one mole of CaCO3 produces one mole of CO2. Therefore, the reaction yields 0.05 moles of CO2.
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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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Thank you very much :) SuperSonic4
 

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