A chemist reacts 50.0 grams of sulfur with 50.0 grams of oxygen to make 83.4 grams of a gas, along with some leftover sulfur. In a different experiment, he reacts 20.0 grams of sulfur with 30.0 grams of oxygen to make 49.9 grams of a gas, along with some leftover oxygen. Are the two gasses he made the same?

1 answer

To determine whether the two gases produced in the two reactions are the same, we first need to analyze the chemical reactions and their stoichiometry.

  1. Reaction of Sulfur and Oxygen: The chemical reaction between sulfur (S) and oxygen (O₂) to form a gas can be represented by the following balanced equation for the formation of sulfur dioxide (SO₂):

    \[ S + O_2 \rightarrow SO_2 \]

    However, sulfur can also react with oxygen to form other oxides like sulfur trioxide (SO₃), depending on the amount of oxygen available.

  2. First Experiment:

    • Reactants: 50.0 g of sulfur (S) + 50.0 g of oxygen (O₂).
    • Product: 83.4 g of gas and leftover sulfur.
  3. Second Experiment:

    • Reactants: 20.0 g of sulfur (S) + 30.0 g of oxygen (O₂).
    • Product: 49.9 g of gas and leftover oxygen.

Next, we need to check the stoichiometry and determine how much sulfur and oxygen were consumed in each reaction.

Step 1: Calculate moles of sulfur and oxygen

  • For Experiment 1:

    • Molar mass of sulfur (S) = 32.07 g/mol
    • Molar mass of oxygen (O₂) = 32.00 g/mol

    \[ \text{Moles of S} = \frac{50.0 \text{ g}}{32.07 \text{ g/mol}} \approx 1.56 \text{ mol} \] \[ \text{Moles of O₂} = \frac{50.0 \text{ g}}{32.00 \text{ g/mol}} \approx 1.56 \text{ mol} \]

  • For Experiment 2:

    \[ \text{Moles of S} = \frac{20.0 \text{ g}}{32.07 \text{ g/mol}} \approx 0.62 \text{ mol} \] \[ \text{Moles of O₂} = \frac{30.0 \text{ g}}{32.00 \text{ g/mol}} \approx 0.94 \text{ mol} \]

Step 2: Understand the reactions and leftovers

  • In the first experiment, we notice that there is leftover sulfur. Hence, not all sulfur reacted, indicating that the oxygen could have potentially formed SO₂ or another gas but the limiting reactant was not oxygen.

  • In the second experiment, there is leftover oxygen, indicating that the sulfur was the limiting reactant here. This implies all sulfur reacted.

Step 3: Identify the products

  1. In Experiment 1, since sulfur is in excess, we cannot finalize which specific gas was produced without further data. However, if the reaction was primarily with available oxygen, likely SO₂ was still produced, but we could be producing mixtures if excess sulfur versus limited oxygen was present.

  2. In Experiment 2, we determine that 20 grams of sulfur reacted entirely given that there was leftover oxygen. This suggests that some fraction of the oxygen was left since we also have to consider the grams of gas produced (49.9 g).

Conclusion:

Given that sulfur reacts with oxygen to produce oxides, the leftover elements and the amount of product produced suggest that both reactions likely yielded the same gas (SO₂), but slight variance in the overall reaction conditions (notably reactant proportions and limitations) implies it must be verified further.

The product formed in each reaction is likely similar (sulfur dioxide, SO₂), but the exact identity may depend on reaction conditions and stoichiometry limits. The difference in leftover materials indicates they could differ in purity, concentration, or other experimental parameters. Thus, from a chemical standpoint, it is not definitive they are the same without further analysis of the gas.