Final answer:
To determine the volume of sulfur dioxide gas produced from burning tetraphosphorus trisulfide, a balanced chemical equation is used followed by molar mass conversion, stoichiometry, and application of the ideal gas law with given conditions.
Step-by-step explanation:
To calculate how many milliliters of sulfur dioxide can be produced from burning 0.241 g of tetraphosphorus trisulfide, we must first write out the balanced chemical equation for the reaction. The reaction is:
4 P4S3 (s) + 21 O2 (g) → 8 P4O10 (s) + 12 SO2 (g)
Next, we'll use the molar mass of P4S3 to convert grams to moles, and then the mole ratio from the balanced equation to determine the moles of SO2 produced.
Finally, we'll use the ideal gas law, PV = nRT, where P is the pressure (converted from torr to atm), V is the volume in liters, n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin. We can rearrange the equation to solve for V (the volume of SO2).
The ideal gas law and the provided conditions (764 torr and 21.4°C, which equals 294.55 K when converted to Kelvin) will enable us to find the answer in liters, which can then be converted to milliliters. Without going through all the calculations in this context, this procedure will yield the volume of SO2 produced.