Consider the following equation:
2SO2(g) + O2(g) → 2SO3(g)
If 50 cm3 of sulphur dioxide react with 60 cm3 of oxygen, what would be the final volume of the gases? ( All volumes are measured at room temperature and pressure. )
A) 35 cm3
B) 50 cm3
C) 85 cm3
D) 100 cm3
---------------------------------------
The answer is C) 85 cm3.
---------------------------------------
**Why?
**Can you show the CLEAR STEPS OF CALCULATION WITH EXPLANATION?
2006-11-04 07:56:26 · 2 個解答 · 發問者 ? 1 in 科學 ➔ 化學
50 cm3 of sulphur dioxide will react with 25 cm3 oxygen and produce 50 cm3 of sulphur trioxide, 60 - 25 = 35 cm3 of oxygen remain unreacted.
Final volume = 0 cm3 SO2 + 35 cm3 O2 + 50 cm3 SO3 = 85 cm3 mixture.
Therefore, Answer C is correct.
2006-11-04 08:21:25 · answer #1 · answered by ? 7 · 0⤊ 0⤋
Base on the equation, two moles SO2 react with 1 mole of oxygen forming two moles of SO3. that means, 50cm3 of SO2 reacts with 25cm3 O2 form 50cm3 of SO3 and 25 cm3 of O2 left. therefore, ans : 50cm3 plus 25cm3 =85cm3.
2006-11-04 09:57:41 · answer #2 · answered by ? 2 · 0⤊ 0⤋