Question

How many grams of sulfur are present in 83.2 grams of
sulfurdioxide?

Answer 1

41.6 g

Step-by-step explanation:

Calculation of the moles of
`SO_2` as:-

Mass = 83.2 g

Molar mass of
`SO_2` = 64.066 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (83.2\ g)/(64.066\ g/mol)`

`Moles_(SO_2)= 1.2987\ mol`

From seen from the formula,

1 mole of sulfur is present in 1 mole of
`SO_2`

So,

1.2987 mole of sulfur is present in 1.2987 mole of
`SO_2`

Moles of sulfur = 1.2987 mol

Molar mass of sulfur = 32.065 g/mol

Mass = Moles*Molar mass =
`1.2987* 32.065\ g` = 41.6 g