Answer:
1.19 g
Step-by-step explanation:
Given that:
Molecular weight of diene
136 g/mol
Molecular weight of Maleic Anhydride
96 g/mol
Mass of crude diene = 2.5 g
Percent of Composition for Peak A = 66%
Percent of Composition for Peak B = 22.66%
Percent of Composition for Peak C = 11.33%
Let us determine the amount of diene in the sample;
So, using
to represent the amount of diene; we have :
=

=

= 1.65 g
Using the limiting reagent equation to determine the amount of Maleic Anhydride

=

=

= 1.19 g
The grams of Maleic Anhydride used to react with 2.5 g sample of crude diene = 1.19 g