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
The average atomic mass of element X is 14.007 u.
The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).
Thus,
0.996 36 × 14.003 u =13.952 03 u
0.003 64 × 15.000 u = <u>0.054 60 u</u>
__________TOTAL = 14.007 u
D, the energy of formation is equal to the dissociation. so from magnesium ions and Hydrogen ions, when they form the lattice energy is equal to formation-dissociation, as they aren't going to dissociate (fixed in lattice) the formation should be the same as the the overall lattice energy.
Answer:
3N2Cl4 + 3CH4--> 3CCl4 + 3N2 + 6H2
This simplifies to:
N2Cl4 + CH4 --> CCl4 + N2 + 2H2
Hey there !
Molar mass HCl :
HCl =1.01 + 35.45 => 36.46 g/mol
Number of moles ( solute) :
n = m / mm
n = 9.63 / 36.46
n = 0.2641 moles
Therefore :
M = moles / volume ( L )
M = 0.2641 / 1.5
=> 0.176 M
