Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
<h3>
Reaction between oxygen and ethene</h3>
Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.
C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O
from the equation above;
3 moles of O₂ ---------> 2(18 g) of water
3.5 moles of O₂ ----------> x
![x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]](https://tex.z-dn.net/?f=x%20%3D%203.2%20%5Ctimes%20%5B%5Cfrac%7B2%20%5C%20moles%20%5C%20H_2O%7D%7B3%20%5C%20moles%20%5C%20O_2%7D%20%20%5D%20%5Ctimes%5B%20%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mole%20%5C%20H_2O%7D%20%5D)
Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
Learn more about reaction of ethene here: brainly.com/question/4282233
#SPJ1
Answer:
ffgghhhhhgffffffcvvvvvvvvvvvvvvvvvv
Explanation:
cccvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Answer:
a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 
⇒ 16
c) No 
was not the limiting reactant.
Explanation:
Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.
a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.
b) ⇒ 16
c) was not the limiting reactant based on the mol to mol ratio of and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of would also be produced.
Less water = less weight to make it rise
More water = more weight to make it dive
Hey there:
Correct answer is :
(b) NaNH₂
Sodium azanide NaNH₂ is the conjugate base of ammonia NH₃
Correct answer is :
(b) NaNH₂
I hope this will help !
