Final answer:
To find the volume of SO2 produced from 50.0 g of FeS, calculate the moles of FeS, use the stoichiometry to find moles of SO2, and then convert to liters at STP. The closest answer to the calculated 12.7 liters is 11.2 liters, suggesting a possible typo in the provided options.
Step-by-step explanation:
To calculate the volume of SO2 produced at STP from the reaction of 50.0 g of iron(II) sulfide, FeS, we first need to use the balanced chemical equation provided: 4 FeS + 7 O2 --> 2 Fe2O3 + 4 SO2. Here are the steps we'll follow:
- Calculate the molar mass of FeS.
- Determine the number of moles of FeS.
- Use the stoichiometry from the balanced equation to find the moles of SO2 produced.
- Convert the moles of SO2 to liters using the ideal gas law at STP (Standard Temperature and Pressure), where 1 mole equals 22.4 liters.
Calculations:
- Molar mass of FeS = 55.85 (Fe) + 32.07 (S) = 87.92 g/mol
- Moles of FeS = 50.0 g / 87.92 g/mol = 0.569 moles
- Moles of SO2 produced = (0.569 moles FeS) (4 moles SO2 / 4 moles FeS) = 0.569 moles SO2
- Volume of SO2 at STP = 0.569 moles × 22.4 liters/mole = 12.7 liters
The nearest answer choice to 12.7 liters is therefore c) 11.2 L, assuming that the answer list provided has a typo because none of the options match exactly.