AES Mix columns bit example
If you look at the code in the Wikipedia we will see this part of the code;


h = (unsigned char)((signed char)r[c] &gt;&gt; 7);
b[c] = r[c] &lt;&lt; 1;
[c] ^= 0x1B &amp; h;


h stores the leftmost bit of c

c is x-or'ed with 0x1b but not with 0x11b  why?

When shifting r left by one, the MSB value is discarded. Before discarding, we hold this value with h. 


If h==0 than modulus operation is not required. The 0x1B &amp; h = 0x0 so there is no x-or with 0x1b
If h=1 than modulus operation is required. The 0x1B &amp; h = 0x1B. The x-or with 0x1b is performed.
Note that: it is not x-or with 0x11b it is with 0x1b since we discarded the MSB 1 while shifting. Therefore, there is no need to x-or with 0x11b, x-oring with 0x1b is enough. Now, it is clear that everthing is in Bytes.


A small example;

Here [ ] represents 8-bit.

Let multiply $a = (x^7+x^6+x^3+x)$ by $\{2\}$


binary representation of  $a = [11001010]$
$h = 1 = [00000001]$
$b = a \ll 1 = 1[10010100]$ but everthing is in Bytes, therefore 
$b = [10010100]$
$h \wedge 0x1b = [00000001] \wedge [00011011] = [00011011] = 0x1b$
$b \oplus 0x1b = [10010100] \oplus [00011011] = [10001111] = 0x8f$