A gas mixture contains SO 2 SO2 ( molar mass = 64 g/mol ) (molar mass=64 g/mol) and no other source of sulfur. If the mixture is 10 % 10% sulfur by mass, what is the percentage (by mass) of SO 2 SO2 in the mixture?

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Alleei Alleei · Answer 1 · 2019-12-11T07:03:14+01:00

Answer : The percentage (by mass) of [tex]SO_2[/tex] in the mixture is 20 %

Explanation :

As we are given that, 10 % sulfur by mass that means 10 grams of sulfur present in 100 grams of mixture.

Mass of sulfur = 10 g

Mass of mixture = 100 g

Now we have to calculate the mass of [tex]SO_2[/tex].

As, 32 grams of sulfur present in 64 grams of [tex]SO_2[/tex]

So, 10 grams of sulfur present in [tex]\frac{10}{32}\times 64=20[/tex] grams of [tex]SO_2[/tex]

Thus, the mass of [tex]SO_2[/tex] is 20 grams.

Now we have to calculate the percentage (by mass) of [tex]SO_2[/tex] in the mixture.

[tex]\% \text{ of }SO_2=\frac{\text{Mass of }SO_2}{\text{Mass of mixture}}\times 100[/tex]

[tex]\% \text{ of }SO_2=\frac{20g}{100g}\times 100=20\%[/tex]

Therefore, the percentage (by mass) of [tex]SO_2[/tex] in the mixture is 20 %