Respuesta :
Answer: The amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]
Explanation:
Converting given amount of mass in tons to grams, we use the conversion factor:
1 ton = 907185 g .......(1)
So, [tex]2.68\times 10^7=2.431\times 10^{13}g[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)
Given mass of sulfur dioxide = [tex]2.431\times 10^{13}g[/tex]
Molar mass of sulfur dioxide = 64 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of sulfur dioxide}=\frac{2.431\times 10^{13}g}{64g/mol}=3.79\times 10^9mol[/tex]
For the given chemical reaction:
[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of sulfur dioxide is produced from 1 mole of sulfur
So, [tex]3.79\times 10^9[/tex] moles of sulfur dioxide will be produced from = [tex]\frac{1}{1}\times 3.79\times 10^9=3.79\times 10^9[/tex] moles of sulfur.
Now, calculating the mass of sulfur using equation 2:
Moles of sulfur = [tex]3.79\times 10^9mol[/tex]
Molar mass of sulfur = 32 g/mol
Putting values in equation 2, we get:
[tex]3.79\times 10^9mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Moles of sulfur}=121.54\times 10^{11}g[/tex]
Converting this value in tons using conversion factor 1, we get:
[tex]\Rightarrow (\frac{1ton}{907185g})\times 121.54\times 10^{11}g\\\\\Rightarrow 13397491.6tons=1.34\times 10^7tons[/tex]
Hence, the amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]
