How many grams of Br is added to 3.5 grams of pentene?

Decision:
1. Let’s write down the reaction equation:
Н3С – СН2 – СН2 – СН = СН2 + Br2 = H3C – CH2 – CH2 – CHBr – CH2Br – addition reaction, dibromopentane is released;
2. Let’s make calculations using the formulas:
M (C5H10) = 12 * 5 + 10 = 70 g / mol;
M (Br2) = 2 * 79.9 = 159.8 g / mol;
3. Let us calculate the number of moles of pentene, if the mass is known:
Y (C5H10) = m / M = 3.5 / 70 = 0.05 mol;
4. Let’s make a proportion according to the reaction equation:
0.05 mol (C5H10) – X mol (Br2);
– 1 mol -1 mol from here, X mol (Br2) = 0.05 * 1/1 = 0.05 mol;
5. Let’s calculate the mass of bromine by the formula:
m (Br2) = Y * M = 0.05 * 159.8 = 7.99 g.
Answer: bromine weighing 7.99 g is required to carry out the reaction.