NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer:
The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.
Explanation:
The given reaction undergoes
mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.
Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.
After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.
When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.
The chemical reaction is as follows.
The element bromine is not a reddish-brown liquid. Liquid is the substance bromine.
M=DV
M=3.103 g/mL * 19.8 mL = 61.44 g