If I'm not mistaken, the answer should be B. Oxidation
Answer:
Percentage yield =58.7 %
Explanation:
moles of calcium carbonate as a reactant =
as per molar equation , 0.2068 moles of calcium carbonate produces 0.2068 moles of CaO .
amount of CaO = 0.2068 mole x (40.078+15.999) g/moles = 11.598 gram
percentage yield = actual yield / theoritical yield = 6.81/ 11.598 = 0.587 or 58.7 %
V₁ = 900.0 mL
T₁ = 27.0ºC + 273 = 300 K
V₂ = ?
T₂ = 132.0 + 273 = 405 K
V₁ / T₁ = V₂ / T₂
900.0 / 300 = V₂ / 405
300 x V₂ = 405 x 900.0
300 x V₂ = 364500
V₂ = 364500 / 300
V₂ = 1215 mL
hope this helps!
I'm not sure but I think is C or D.