Question

Please help me find the final answer for number six

Answer 1

1.28x10^23 molecules of carbon dioxide are expelled.

1st) It is necessary to write the balanced equation of combustion from sugar:

`C_(12)H_(22)O_(11)+12O_2\text{ }\rightarrow12CO_2+11H_2O`

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:

`\begin{gathered} 342gC_(12)H_(22)O_(11)-528gCO_2 \\ 6.06gC_(12)H_(22)O_(11)-x=(6.06g\cdot528g)/(342g) \\ x=9.36gCO_2 \end{gathered}`

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:

`\begin{gathered} 44gCO_2-6.022\cdot10^(23)molecules \\ 9.36gCO_2-x=(9.36gCO_2\cdot6.022\cdot10^(23)molecules)/(44gCO_2) \\ x=1.28\cdot10^(23)molecules \end{gathered}`

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:

`6.06g\cdot(528g)/(342g)\cdot\frac{6.022\cdot10^(23)\text{molecules}}{44g}=1.28\cdot10^(23)molecules`

And the result will be the same.