In the formation of so2 and so3 the ratio of their weights of oxygen which combines with 10kg of sulphur is
Answers
The ratio of weights of Oxygen = 2 : 3
- Given , formation of SO2 and SO3
- Mass of Sulphur = 10 kg
- Moles of Sulphur = 10 * 10^3 g /32 g/mol = 0.3125 * 10^3 moles
- Now the reactions involved are: S + O2 ⇒ SO2 , 2S + 3 O2 ⇒ 2SO3
- Now ,it is clear 1 mole of S requires 1 mole of O2 to form 1 mole SO2
- Also, 2 moles of S requires 3 moles of O2 to form 2 moles of SO3
- So,0.3125 * 10 ^3 moles of Sulphur combines with 0.3125 * 10^3 moles of Oxygen to from SO2.
- Also, 0.3125 * 10 ^3 moles of S combines with 0.3125 * 10^3 * 3/2 moles of Oxygen to form SO3
- We know, 1 mole of O2 = 32 gm of O2
- 0.3125 * 10 ^3 mole of O2 = 10 * 10^3 gm of O2
- 0.3125 * 3/2 * 10^3 mole of O2 = 15 * 10^3 gm of O2
- Therfore mass ratio of SO2 : SO3 = 10 : 15 = 2:3.
Answer:
2:3
Explanation:
The ratio of weights of Oxygen = 2:3
Given , formation of SO2 and SO3 • Mass of Sulphur = 10 kg
• Moles of Sulphur = 10 * 10^3 g /32 g/mol = = 0.3125 * 10^3 moles
• Now the reactions involved are:
S+ 02 = SO2
2S +3 02 = 2SO3
• Now ,it is clear 1 mole of S requires 1 mole
of 02 to form 1 mole SO2
• Also, 2 moles of S requires 3 moles of O2 to form 2 moles of SO3 S0,0.3125 * 10^3 moles of Sulphur
combines with 0.3125 * 10^3 moles of
Oxygen to from SO2. Also, 0.3125 * 10 13 moles of S combines with 0.3125 * 10^3 * 3/2 moles of Oxygen to form SO3
• We know, 1 mole of 02 = 32 gm of O2 • 0.3125 * 10^3 mole of 02 = 10 * 10^3 gm of 02
0.3125 * 3/2 * 10^3 mole of 02 = 15 * 10^3 gm of 02
Therfore mass ratio of SO2: SO3 = 10:15 = 2:3.