What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.

The amount of product produced during a reaction is known as yield. The term "100% yield" refers to the conversion of the complete reactant into the product.

Consider the following reaction:

\[{\rm{Propene}}+{\rm{HBr}}\to2{\rm{-bromopropane}}\]

Thus, the number of moles of propene is equal to the number of moles of 2-bromopropane.

This reaction means that one mole of propene produces one mole of 2 bromopropane when it reacts with HBr.

The following formula is used to compute the amount of moles of 2-bromopropane produced

Mass of propene = 25.5g

Molar mass of propene (C₃H₆) = 42.08g mol^-1

\begin{aligned}Number\,of\,moles\,of\,propene{\bf{=}}\frac{{{\bf{mass}}}}{{{\bf{molarmass}}}}\\{\bf{=}}\frac{{{\bf{25}}{\bf{.5}}\,{\bf{g}}}}{{{\bf{42}}{\bf{.08}}\,{\bf{g/mol}}}}\\{\bf{=0}}{\bf{.606}}\,{\bf{mol}}\end{aligned}

If the yield is 100 percent, the number of moles of 2-bromopropane produced equals the number of moles of propene consumed. As a result, 0.606 moles of 2-bromopropane are generated.

The mass of 0.606 mol of 2-bromopropane is:

\begin{aligned}{m_{2-bromopropane}}={n_{2-bromopropane}}\times{M_{2-bromopropane}}\\=0.606\,mol\times123\,g/mol\\=74.5\,g\end{aligned}