What is the theoretical yield of ethyl chloride in the reaction of 27.17 g of ethylene with 58.33 g of hydrogen chloride? (for ethylene, mw=28.0 amu; for hydrogen chloride, mw=36.5 amu; for ethyl chloride, mw=64.5 amu.) report your answer to the nearest hundredth of a gram without units.

h2c=ch2 + hcl → ch3ch2cl

63.83.

Explanation:

1.0 mole of ethylene reacts with 1.0 mol of HCl to produce ethyl chloride.

no. of moles of ethylene = mass/molar mass = (27.71 g)/(28.0 g/mol) = 0.9896 mol.

no. of moles of HCl = mass/molar mass = (58.33 g)/(36.5 g/mol) = 1.589 mol.

Now, we need to find the no. of moles of produced ethyl chloride:

∵ 1.0 mole of ethylene produce → 1.0 mole of CH₃CH₂Cl.

∴ 0.9896 mole of ethylene produce → 0.9896 mole of CH₃CH₂Cl.

The mass produces of ethyl chloride = no. of moles x molar mass = (0.9896 mol)(64.5 g/mol) = 63.83 g.

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."