"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

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole