What volume of SO2 is produced when 2.35 g of sulfur burns? (All volumes are measured at STP) S (s) + O2(g) →SO2(g)

1.64 L of SO₂ is produced when 2.35 g of Sulfur burns.

Explanation:

First of all, we need to write the balanced equation as,

S (s) + O₂ (g) → SO₂ (g)

Now we have to find the number of moles of sulfur by using its given mass and molar mass.

1 mol of Sulfur atoms = 32 g of sulfur

Now the given mass is 2.35 g.

So

1 mol of S produces 1 mol of SO₂.

So 0.073 mol of S produces 0.073 mol of SO₂.

At STP, 1 mol of SO₂ occupies a volume of 22.4 L.

0.073 mol of SO₂ occupies a volume of 22.4 × 0.073 = 1.64 L

So 1.64 L of SO₂ is produced.

i just learnt this recently but i need to look through my notes again