Question

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 747 torr and 23.8°C, can be produced from burning 0.576 g of tetraphosphorus trisulfide?

Answer 1

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

To solve this problem we need to use PV=nRT, so first let's convert the given units:

• 23.8 °C → 23.8 + 273.15 = 296.95 K

• 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * $\frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}}$ = 7.85 * 10⁻³ mol SO₂ = n

• Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

• Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL