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Zorluk: OrtaTetraoxosulfate(VI) Acid: Contact Process and Properties
During the first stage of the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, pure sulfur is burned in dry air to produce sulfur(IV) oxide gas according to the equation:
S(s)+O2(g)SO2(g)S(s) + O_2(g) \rightarrow SO_2(g)
What volume of sulfur(IV) oxide gas, in dm3\text{dm}^3 measured at standard temperature and pressure (STP), is produced by the complete combustion of 16.0 g16.0\text{ g} of sulfur?
[Relative atomic mass: S=32S = 32; Molar volume of gas at STP = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1}]

Cevap: 11.2 dm³

Cevap

The volume of sulfur(IV) oxide gas produced at STP is 11.2 dm311.2\text{ dm}^3.
According to the balanced equation S(s)+O2(g)SO2(g)S(s) + O_2(g) \rightarrow SO_2(g), 1 mol1\text{ mol} (32 g32\text{ g}) of sulfur yields 1 mol1\text{ mol} (22.4 dm322.4\text{ dm}^3 at STP) of sulfur(IV) oxide gas. Therefore, 16.0 g16.0\text{ g} of sulfur corresponds to 16.032=0.50 mol\frac{16.0}{32} = 0.50\text{ mol}, which produces 0.50×22.4 dm3=11.2 dm30.50 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3 of SO2SO_2 gas at STP.

Adım Adım Çözüm

1
Calculate the amount in moles of sulfur reacted.
0.50 mol0.50\text{ mol} of sulfur.
Using the formula moles=massmolar mass=16.0 g32.0 g mol1=0.50 mol\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.0\text{ g}}{32.0\text{ g mol}^{-1}} = 0.50\text{ mol}.
2
Use the mole ratio from the balanced chemical equation to find moles of sulfur(IV) oxide gas formed.
0.50 mol0.50\text{ mol} of SO2(g)SO_2(g).
The equation shows a 1:1 stoichiometric ratio between S(s)S(s) and SO2(g)SO_2(g).
3
Calculate the gas volume at standard temperature and pressure (STP).
11.2 dm311.2\text{ dm}^3.
Multiply the moles of gas by the molar volume at STP: V=0.50 mol×22.4 dm3 mol1=11.2 dm3V = 0.50\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3.

Anahtar Kavram

Stoichiometric Volume Calculations for Gas Generation in the Contact Process
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