1.28x10^23 molecules of carbon dioxide are expelled.
1st) It is necessary to write the balanced equation of combustion from sugar:
[tex]C_{12}H_{22}O_{11}+12O_2\text{ }\rightarrow12CO_2+11H_2O[/tex]2nd) We need to look for the molar mass of C12H22O11 and CO2 to make a relation between those values:
- C12H22O11 molar mass: 342 g/mol
- CO2 molar mass: 44 g/mol
3rd) According to the balanced chemical equation, 1 mol of C12H22O11 produces 12 moles of CO2. Using the molar mass of the compound, in grams, 342g of C12H22O11 produces 528 g (44gx12) of CO2.
Now, with a mathematical Rule of Three we can find the amount of CO2 that is produced from 6.06g of C12H22O11:
[tex]\begin{gathered} 342gC_{12}H_{22}O_{11}-528gCO_2 \\ 6.06gC_{12}H_{22}O_{11}-x=\frac{6.06g\cdot528g}{342g} \\ x=9.36gCO_2 \end{gathered}[/tex]4th) Knowing that 6.06g of jolly rancher produces 9.36g of carbon dioxide, and using the Avogadro's number (6.022x10^23 molecules/mol), we can find the molecules of CO2:
[tex]\begin{gathered} 44gCO_2-6.022\cdot10^{23}molecules \\ 9.36gCO_2-x=\frac{9.36gCO_2\cdot6.022\cdot10^{23}molecules}{44gCO_2} \\ x=1.28\cdot10^{23}molecules \end{gathered}[/tex]So, 1.28x10^23 molecules of carbon dioxide are expelled from the combustion of one jolly rancher.
We can write it in just one equation like this:
[tex]6.06g\cdot\frac{528g}{342g}\cdot\frac{6.022\cdot10^{23}\text{molecules}}{44g}=1.28\cdot10^{23}molecules[/tex]And the result will be the same.