Answer:
Explanation:
From the statement of the problem,
B₂S₃
+ H₂O
→ H₃BO₃
+ H₂S
B₂S₃ + H₂O → H₃BO₃ + H₂S
We that the above expression does not conform with the law of conservation of mass:
To obey the law, we need to derive a balanced reaction equation:
Let us use the mathematical method to obtain a balanced equation.
let the balanced equation be:
aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S
where a, b, c and d will make the equation balanced.
Conservating B: 2a = c
S: 3a = d
H: 2b = 3c + 2d
O: b = 3c
if a = 1,
c = 2,
b = 6,
2d = 2(6) - 3(2) = 6, d = 3
Now we can input this into our equation:
B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S
B₂S₃
+ 6H₂O
→ 2H₃BO₃
+ 3H₂S
<h2>
Answer</h2>
Bromination:
Any reaction or process in which bromine (and no other elements) are introduced into a molecule.
Bromonium Ion:
The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)
Mechanism:
Step 1:
In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".
Step 2:
When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.
In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.
We are given with
136 g P4
excess oxygen
The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles
Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5