A sample of 4.20 g of propene is treated with excess hydrogen bromide gas in the absence of a…

Step 1. Propene. $\dfrac{4.20\ \text{g}}{42.0\ \text{g/mol}} = 0.100\ \text{mol}$.

Step 2. Hydrogen bromide. The ratio is 1 : 1, so $0.100\ \text{mol}$ is needed: $0.100\ \text{mol} \times 22.4\ \text{L/mol} = 2.24\ \text{L}$.

Step 3. Product. Hydrogen adds to the carbon that already has more hydrogen atoms, and bromine goes to the middle carbon. The product is 2-bromopropane. Theoretical mass: $0.100\ \text{mol} \times 123.0\ \text{g/mol} = 12.3\ \text{g}$. At $80\ \%$: $12.3\ \text{g} \times 0.80 = 9.84\ \text{g}$.