What mass of sulfur must be used to pro- duce 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.

                     25.05 g of S₈

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

girl imma just put my comment here because i need a couple point okurrttt