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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 ?

    Many Thanks

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    Quote Originally Posted by mathlearn View Post
    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 ?

    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 $ \displaystyle n = \dfrac{m}{M_r} \text{ or } \text{moles} = \dfrac{\text{mass}}{\text{molar mass}}$

    $ \displaystyle n_{CaCO_3} = \dfrac{5}{100} = \dfrac{1}{20} = 0.05 \text{mol}$


    Next consider the balanced reaction that is taking place here which is that of a carbonate with acid

    $ \displaystyle CaCO_{3 (s)} + 2HCl_{(aq)} \rightarrow CaCl_{2 (s)} + CO_{2 (g)} + H_2O_{(l)}$

    From the equation above we see that one mole of $ \displaystyle CaCO_3$ produces one mole of $ \displaystyle CO_2$ which means you'll have $ \displaystyle 0.05$ moles of $ \displaystyle CO_2$ which is the answer to your question.

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