Question

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

Answer 1

Answer:

41.6 g

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{83.2\ g}{64.066\ g/mol}$

$Moles_{SO_2}= 1.2987\ mol$

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 = $1.2987\times 32.065\ g$ = 41.6 g