"Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfur dioxide gas. How many milliliters of sulfur dioxide, measured at 751 torr and 21.0°C, can be produced from burning 0.869 g of tetraphosphorus trisulfide?"

Can be produced 288mL of SO₂

Explanation:

Based in the reaction:

P₄S₃ + 8O₂ → P₄O₁₀ + 3SO₂

Where 1 mole of tetraphosphorus trisulfide reacts producing 3 moles of sulfur dioxide gas.

0.869g of tetraphosphorus trisulfide (Molar mass of P₄S₃: 220.09g/mol) are:

0.869g P₄S₃ ₓ (1mol / 220.09g) = 3.948x10⁻³ moles of P₄S₃

As 3 moles of SO₂ are produced per mole of P₄S₃:

3.948x10⁻³ moles of P₄S₃ ₓ (3 moles SO₂ / 1 mole P₄S₃) = 0.0118 moles SO₂

Using PV = nRT

V = nRT / P

Where n are 0.0118 moles, R gas constant (0.082atmL/molK), T absolute temperature (21.0°C + 273.15K = 294.15K), and P pressure (751torr / 760 = 0.988atm).

Replacing:

V = 0.0118molₓ0.082atmL/molKₓ294.15K / 0.988atm

V = 0.288L = 288mL

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be