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Answers
12. Balanced Reaction of this Process is :
CaCO₃ === On Heating ===> CaO + CO₂
⇒ 100 grams of CaCO₃ on heating gives 56 grams of CaO and 44 grams of CO₂
⇒ 50 grams of CaCO₃ on heating gives 28 grams of CaO and 22 grams of CO₂
Given that on heating 60 grams of Lime stone Produced 22 grams of CO₂
⇒ As the Reaction resulted in 22 grams of CO₂ and we know that from Balanced Equation 22 grams of CO₂ requires 50 grams of CaCO₃ - We can Clearly come to a Conclusion that 60 grams of Lime stone contains 50 grams of CaCO₃ and 10 grams of Some other Chemical substance.
⇒ The % of CaCO₃ in the Limestone = (50/60) × 100 = 83.33%
Option - (3) is the Answer
13. Balanced Reaction for this Process is :
S + O₂ === On Burning ===> SO₂
⇒ 32 grams of Sulphur on burning with 32 grams of Oxygen Produces 64 grams of Sulphur dioxide
⇒ 16 grams of Sulphur on burning with 16 grams of Oxygen Produces 32 grams of Sulphur dioxide
At STP : 22.4 liters of Sulphur Dioxide weighs 64 grams
⇒ At STP : 11.2 liters of Sulphur Dioxide weighs 32 grams
Given that 20 grams of Sulphur on burning in air produced 11.2 liters of SO₂
The Given Question changes to :
20 grams of Sulphur on burning in air produced 32 grams of SO₂
But from the Balanced Reaction 32 grams of SO₂ is produced only with 16 grams of Sulphur.
By this We can conclude that in 20 grams of Sulphur only 16 grams is burned in Oxygen and remaining 20 - 16 = 4 grams of sulphur is Unreacted.
⇒ The % of Unreacted Sulphur = (4/20) × 100 = 20%
Option - (2) is the Answer
Answer:
(12)3
(13) 2 is the right answer