jmc

We begin by multiplying the units digit: $8_9\times 5_9=40_{10}=44_9$. So, we write down a $4$ and carry-over another $4$. Moving on to the next digit, we need to evaluate $1_9\times 5_9+4_9=9_{10}=10_9$. Thus, the next digit is a $0$ and a $1$ is carried over. Finally, the leftmost digits are given by the operation $2_9\times 5_9+1_9=11_{10}=12_9$. Writing this out, we have \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c} & & & & \stackrel{1}{2} & \stackrel{4}{1} & \stackrel{}{8}_9 \\ & & & \times & & & 5_9 \\ \cline{4-7} & & & 1 & 2 & 0 & 4_9 \\ \end{array}So our final answer is $\boxed{1204_9}$.