The answer is
decomposition
Answer:
67.5% ≅ 67.6%
Explanation:
Given data:
Mass of water = 17.0 g
Mass of oxygen produced (actual yield)= 10.2 g
Percent yield of oxygen = ?
Solution:
Chemical equation:
2H₂O → 2H₂ + O₂
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 17.0 g/ 18.016 g/mol
Number of moles = 0.944 mol
Now we will compare the moles of oxygen with water to know the theoretical yield of oxygen.
H₂O : O₂
2 : 1
0.944 : 1/2×0.944 = 0.472 mol
Mass of oxygen:
Mass = number of moles× molar mass
Mass = 0.472 mol × 32 g/mol
Mass = 15.104 g
Percent yield:
Percent yield = [Actual yield / theoretical yield] × 100
Percent yield = [ 10.2 g/ 15.104 g] × 100
Percent yield = 0.675 × 100
Percent yield = 67.5%
Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
I think B sorry if I’m wrong
Explanation
