Answer
a) The modern catalytic method.
b) $55.97\%$
Work Step by Step
a) R1
Reactants: $C_2H_{6}O_2Cl_2Ca, 173.05\ g/mol$
Product: $44.05\ g/mol$
Atom economy: $44.05/173.05\times 100\%=25.46\%$
R2
Reactants: $C_2H_{4}O, 44.05\ g/mol$
Product: $44.05\ g/mol$
Atom economy: $44.05/44.05\times 100\%=100.0\%$
The modern catalytic method.
b) Number of moles of ethylene:
$867\ g\div 28.045\ g/mol=30.905\ mol$
Mass of oxyde:
$30.905\ mol\times 44.05\ g/mol=1361.4\ g$
Percent yield:
$762/1361.4\times100\%=55.97\%$
