Calculations with angular momentum Is the following correct, when adding 3 angular momenta/spins:
\begin{align}
1\otimes 1\otimes \frac{1}{2}&=\left(1\otimes 1\right)\otimes \frac{1}{2}
\\
&=\left(2\oplus 1\oplus 0\right)\otimes \frac{1}{2}
\\
&=\left(2\otimes \frac{1}{2}\right)\oplus \left(1\otimes \frac{1}{2}\right)\oplus \left(0\otimes \frac{1}{2}\right)
\\
&=\left(\frac{5}{2}\oplus\frac{3}{2}\right)\oplus \left(\frac{3}{2} \oplus\frac{1}{2}\right)\oplus \left(\frac{1}{2}\right)
\\
&=\frac{5}{2}\oplus\frac{3}{2}\oplus\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{1}{2}.
\end{align}
Comments:


*

*$\oplus$ and $\otimes$ are commutative.

*$0\otimes\frac{1}{2}$ makes $0$ disappear on the line before the last one... I don't understand that. It also happens with $0\otimes 1$, but that feels wrong. Can I not get a total spin of 0 in that case?

 A: $\newcommand{\ket}[1]{\left| #1 \right>}$
Note that $ l=0$ has only one state $m=0$. Therefore the tensor product of $l=1$ and $l=0$ can be written as:
$$
(l=1)\otimes (l=0) = \left\{
\begin{array}
&\ket{l= 1,m=1} \otimes \ket{l=0,m=0} \\
\ket{l=1,m=1} \otimes \ket{l=0,m=0} \\
\ket{l=1,m=1} \otimes \ket{l=0,m=0} \\
\end{array} \right\}=(l=1)
$$
As you have probably noticed this is just the spectrum of $l=1$ state hence you just consider it as $l=1$ triplet.
Edit:
For some reason thought you asked $(l=1) \otimes (l=0)$ but you can equally well apply it to $ \left(l= \frac{1}{2} \right) \otimes (l=0)$
A: Your final result looks right to me. Everything should be half-integers.
A basic rule of combining two quantized angular momenta is that the quantum number of the resultant can be anywhere between the sum of the original quantum numbers and the absolute value of the difference of them, in integer steps. Consider $\ell_1$ = 1 combining with $\ell_2$=3. The quantum number, $L$, of the combination can be between |3-1|=2 and 3+1=4, stepping by 1, or $L = 2, 3, \text{ or } 4$.
So if you have only one particle with a half-integer quantum number, the combination must be a half-integer quantum number. There is no way to get 0.
Also, with quantum numbers of 1 and 0, the result would be between |1-0| and 1+0, which is 1.
The reason for this rule is based in the mathematics of preserving angular momentum wave functions. If you study more QM, you'll eventually get to rotation group theory, 3j symbols, Clebsch-Gordon coefficients, etc, which will all help explain the rule. But the rule works.
A: To dramatize the point of the above correct answers, let me rewrite your multiplets the way mathematicians, or SU(3), SU(5),...,  mavens, write them, through the actual dimensionality 2l+1 of the multiplets, in boldface.
So then, l=1/2 is a doublet, 2; 1 is a triplet, 3; 3/2 is a quartet, 4; and 5/2 a sextet, 6; and 0 a singlet, 1, the source of your confusion. 
Then your Clebsch-Gordan series for the Kronecker product is expressible 
as just
 $$
{\bf 3 } \otimes {\bf 3 } \otimes {\bf 2 }= {\bf 6 } \oplus {\bf 4 } \oplus {\bf 4 } \oplus {\bf 2 } \oplus {\bf 2 } ~, 
$$
which manifestly matches at the level of elementary school arithmetic the dimensionality of the vector space of the entire reducible representation. 
