Can u pls help me find the volume of oxygen gas that reacts with 10 dm3 of sulfur dioxide gas measured at the same temperature and pressure. 2SO2(g) + O2(g) 2SO3(g) Thank youuu
Answers
Answer: 5dm³
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mol(2SO2) = v/24 = 10(dm^3)/24 = 5/12mol
ratio of SO2:O2 = 2:1
mol(O2) = 5/12mol ÷ 2 = 5/24mol
v(O2) = 24 × 5/24mol = 5dm³
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Explanation: You are told to find the volume of oxygen gas that reacts with 10 dm³ of sulfur dioxide gas measured at the same temperature and pressure, and the equation for the reaction is: 2SO2(g) + O2(g) --> 2SO3(g).
A key equation here is volume(dm³) = 24 × moles.
First, you find the number of moles of sulphur dioxide (2SO2):
mol(2SO2) = v/24 = 10dm³/24 = 5/12mol
You know that the ratio of 2SO2:O2 is 2:1, so the ratio of the number of moles of sulphur dioxide to oxygen = 2:1.
This means that the number of moles of oxygen (O2) will equal to half of the number of moles of sulphur dioxide (2SO2), which is 5/12mol:
mol(O2) = 5/12mol ÷ 2 = 5/24mol
Knowing this, you can now work out the volume of oxygen by using the formula: volume(dm³) = 24 × moles.
Therefore, v(O2) = 24 × 5/24mol = 5dm³.