Question

I really need help! Please help me out! The picture is below

Answer 1

• The mass of DDT is 1026.6 g
• Chlorobenze is in excess while chloral is limiting
• Mass of excess chlorobenzene is 742.5 g
• The percent yield is 21%.

Number of moles of chlorobenzene = 1164 g/113 g/mol

= 10.3 moles

Number of moles of chloral = 474 g/165 g/mol

= 2.9 moles

If 2 moles of chlorobenzene reacts with 1 moles of chloral

10.3 moles of chlorobenzene reacts with 10.3 * 1/2

= 5.15 moles

The chloral is the limiting reactant

If 1 mole of chloral produces 1 mole of DDT

2.9 moles of chloral produces 2.9 moles of DDT

Mass of the DDT produced = 2.9 moles * 354 g/mol

= 1026.6 g

If 1 mole of chloral reacts with 2 moles of chlorobenzene

2.9 moles of chloral reacts with 2.9 * 2/1

= 5.8 moles

Excess moles of chlorobenzene = 10.3 moles - 5.8 moles

= 4.5 moles

Mass of excess chlorobenzene = 4.5 moles * 165 g/mol

= 742.5 g

Percent yield = 216 g/1026.6 g * 100/1

= 21%