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Difficulty: HardSulfur Allotropes, Hydrogen Sulfide, and Sulfur(IV) Oxide

A sample of 4.80 g4.80\text{ g} of pure rhombic sulfur is completely burned in excess oxygen gas at room temperature and pressure (RTP). What is the volume of sulfur(IV) oxide gas liberated at RTP, and what is the change in the oxidation state of sulfur during this combustion process?

(Relative atomic mass: S=32.0S = 32.0; Molar volume of gas at RTP = 24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1})

  1. 3.60 dm33.60\text{ dm}^3 and an increase from 0 to +4+4Answer
  2. B
    3.36 dm33.36\text{ dm}^3 and an increase from 0 to +4+4
  3. C
    3.60 dm33.60\text{ dm}^3 and an increase from 0 to +6+6
  4. D
    1.80 dm31.80\text{ dm}^3 and an increase from −2-2 to +4+4

Answer

The volume of sulfur(IV) oxide gas produced at RTP is 3.60 dm33.60\text{ dm}^3, and the oxidation state of sulfur increases from 0 to +4+4.
Burning 4.80 g4.80\text{ g} (0.15 mol0.15\text{ mol}) of sulfur produces 0.15 mol0.15\text{ mol} of SO2SO_2 gas. Multiplying by the molar volume at RTP (24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}) gives 3.60 dm33.60\text{ dm}^3. In elemental rhombic sulfur, sulfur has an oxidation state of 0, which increases to +4+4 in SO2SO_2.

Step-by-Step Solution

1
Calculate the amount of sulfur reacted in moles.
Moles of S=4.80 g32.0 g mol−1=0.15 mol\text{Moles of } S = \frac{4.80\text{ g}}{32.0\text{ g mol}^{-1}} = 0.15\text{ mol}
Dividing the mass of sulfur by its relative atomic mass gives the mole amount.
2
Write the balanced chemical equation and determine the mole ratio.
S(s)+O2(g)→SO2(g)S(s) + O_2(g) \rightarrow SO_2(g) (Mole ratio of S:SO2=1:1S : SO_2 = 1 : 1, so 0.15 mol0.15\text{ mol} of SO2SO_2 is formed).
Direct combustion of sulfur yields sulfur(IV) oxide.
3
Calculate the volume of SO2SO_2 gas produced at RTP.
Volume=0.15 mol×24.0 dm3 mol−1=3.60 dm3\text{Volume} = 0.15\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 3.60\text{ dm}^3
At room temperature and pressure (RTP), 1 mol1\text{ mol} of any gas occupies 24.0 dm324.0\text{ dm}^3.
4
Determine the change in oxidation state of sulfur.
Elemental sulfur S(s)S(s) has an oxidation number of 0. In SO2SO_2, oxygen is −2-2, so x+2(−2)=0  ⟹  x=+4x + 2(-2) = 0 \implies x = +4. The change is from 0 to +4+4.
Free uncombined elements have an oxidation state of 0.

Key Concept

Stoichiometry of gas liberation at RTP and oxidation state changes during non-metal combustion
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Topics

Sulfur Allotropes, Hydrogen Sulfide, and Sulfur(IV) Oxide

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