How many grams of hydrogen bromide is required to obtain 18.8 g of bromoethane from acetylene.

How many grams of hydrogen bromide is required to obtain 18.8 g of bromoethane from acetylene. 21 g of bromoethane from ethylene?

Problem 1
Given:
m (C2H3Br) = 18.8 g

To find:
m (HBr) -?

1) C2H2 + HBr => C2H3Br;
2) n (C2H3Br) = m (C2H3Br) / M (C2H3Br) = 18.8 / 107 = 0.18 mol;
3) n (HBr) = n (C2H3Br) = 0.18 mol;
4) m (HBr) = n (HBr) * M (HBr) = 0.18 * 81 = 14.6 g.

Answer: The HBr mass is 14.6 g.

Problem 2
Given:
m (C2H5Br) = 21 g

To find:
m (HBr) -?

1) C2H4 + HBr => C2H5Br;
2) n (C2H5Br) = m (C2H5Br) / M (C2H5Br) = 21/109 = 0.19 mol;
3) n (HBr) = n (C2H5Br) = 0.19 mol;
4) m (HBr) = n (HBr) * M (HBr) = 0.19 * 81 = 15.4 g.

Answer: HBr weighs 15.4 g.