Question

What is the theoretical yield of ethyl chloride in the reaction of 19.2 g of ethylene with 52 g of hydrogen chloride? (For ethylene, MW=28.0amu; for hydrogen chloride, MW=36.5amu; for ethyl chloride, MW=64.5amu.)
H2C=CH2+HCl→CH3CH2Cl

Answer 1

$\frac{(19.2 \ g \ H_2C=CH_2)}{(28.0\ g\ H_2C=CH_2/mol)}$$= 0.68571 \ mol \ H_2C=CH_2$

$\frac{(52 \ g\ HCl)}{(36.5 g HCl/mol)}$ $= 1.4247 \ mol\ HCl$

$0.68571$ mole of $H_2C=CH_2$ would react completely with $0.68571$ mole of $HCl$ , but there is more $HCl$ present than that, so $HCl$ is in excess and $H_2C=CH_2$ is the limiting reactant.

The reaction is :

$H_2C=CH_2+HCl→CH_3CH_2Cl$

Moles of ethene$=$ Moles of $CH_3CH_2Cl$ formed.

$(0.68571\ mol\ H_2C=CH_2)$ $\times$ $(64.5 \ g \ CH_3CH_2Cl/mol) =$

$44.2\ g\ CH_3CH_2Cl$ in theory.