Answer:
2 mole of C4H10 were used in this reaction.
my answer would be
fluorine, chlorine, oxygen
The grams of potassium chlorate that are required to produce 160 g of oxygen is 408.29 grams
<u><em>calculation</em></u>
2 KClO₃→ 2 KCl + 3O₂
Step 1: find the moles of O₂
moles = mass÷ molar mass
from periodic table the molar mass of O₂ = 16 x2 = 32 g/mol
moles = 160 g÷ 32 g/mol = 5 moles
Step2 : use the mole ratio to determine the moles of KClO₃
from equation given KClO₃ : O₂ is 2:3
therefore the v moles of KClO₃ = 5 moles x 2/3 = 3.333 moles
Step 3: find the mass of KClO₃
mass= moles x molar mass
from periodic table the molar mass of KClO₃
= 39 + 35.5 + (16 x3) =122.5 g/mol
mass = 3.333 moles x 122.5 g/mol =408.29 grams
Hydroboration-oxidation is a reaction which follow anti markonikov orientation.
The stereochemistry of hydroboration-oxidation follows syn-adition being a stereospecific reaction. That means that boron and hydrogen are added from the same side. Then, when the oxidation occurs, the boron is replaced by an hydroxyl having the same geometric position.