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

#### mathlearn

##### Active member

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

#### SuperSonic4

##### Well-known member
MHB Math Helper
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.