Chapter 3 - Chemical Compounds - Example 3-3 - Using Relationships Derived from a Chemical Formula - Page 77: Practice Example A

18.93 g Br.

Grams of Br= $25.00\,mL\,C_{2}HBrClF_{3}\times\frac{1.871\,g\,C_{2}HBrClF_{3}}{1\,mL\,C_{2}HBrClF_{3}}\times\frac{1\,mol\,C_{2}HBrClF_{3}}{197.4\,g\,C_{2}HBrClF_{3}}\times\frac{1\,mol\,Br}{1\,mol\,C_{2}HBrClF_{3}}\times\frac{79.904\,g\,Br}{1\,mol\,Br}$ $=18.93\,g\,Br$