Question

What mass of sulfur must be used to produce 17.5 L of gaseous sulfur dioxide at STP according to the following equation?

S8 (s) + 8 O2 (g) −→ 8 SO2 (g)
Answer in units of g.

Answer 1

25.05 g of S₈

Step-by-step explanation:

The balance chemical equation for given synthetic reaction is,

S₈ + 8 O₂ → 8 SO₂

Step 1: Calculate Moles of SO₂ at STP:

According to Avogadro's Law, at STP (0 ° C and 1 atm) every ideal gas occupies 22.4 L of volume. Therefore, the moles contained by 17.5 L of SO₂ are as;

Moles = 17.5 L / 22.4 L

Moles = 0.781 moles of SO₂

Step 2: Calculate Moles of S₈ consumed:

According to balance equation,

8 moles of SO₂ were produced by = 1 mole of S₈

So,

0.781 moles of SO₂ will be produced by = X moles of S₈

Solving for X,

X = 0.781 mol × 1 mol / 8 mol

X = 0.0976 moles of S₈

Step 3: Calculate Mass of S₈ as;

Mass = Moles × M.Mass

Mass = 0.0976 mol × 256.52 g/mol

Mass = 25.05 g of S₈