Encryption and HMAC in a PSKC file This is the PSKC standard

https://tools.ietf.org/rfc/rfc6030.txt

In Section 6.1

  6.1.  Encryption Based on Pre-Shared Keys


They have the following

Plain Text = 3132333435363738393031323334353637383930
Algo = aes-128-cbc 
pkcs#5 padding
Encryption Key = 12345678901234567890123456789012
IV = 000102030405060708090a0b0c0d0e0f
Encrypted text = ESIzRFVmd4iZABEiM0RVZgKn6WjLaTC1sbeBMSvIhRejN9vJa2BOlSaMrR7I5wSX


How do they get this? I tried using openssl command line

echo 3132333435363738393031323334353637383930 | openssl enc -aes-128-cbc  -nosalt -iv 000102030405060708090a0b0c0d0e0f -a -pass pass:12345678901234567890123456789012 


I get the following

HgFUJvBN9ztHcsIXNjvoTJJfOsoIq7XlOpdYCn+nk2TxQFrOJGhN+dYOHpFuIBQn


Likewise for the hmac-sha1

key = 1122334455667788990011223344556677889900
hmac-sha1 = Su+NvtQfmvfJzF6bmQiJqoLRExc=


I do it with openssl command line

echo  3132333435363738393031323334353637383930 | openssl sha1 -hmac 1122334455667788990011223344556677889900 | openssl enc -base64

(stdin)= eb6093578fdb69477f9a3322dcda28ce28b8aa1e

echo eb6093578fdb69477f9a3322dcda28ce28b8aa1e  | openssl enc -a -e 


I get 

ZWI2MDkzNTc4ZmRiNjk0NzdmOWEzMzIyZGNkYTI4Y2UyOGI4YWExZSAgDQo=


What am I doing wrong?
0: you have the wrong CipherValue. The one you show is in MACMethod/MACKey and is the encryption 
of the MAC key, see 6.1.1. The encryption of the subject key is in Key/Data/Secret/EncryptedValue 
and is AAECAwQFBgcICQoLDA0OD+cIHItlB3Wra1DUpxVvOx2lef1VmNPCMl8jwZqIUqGv.

1: openssl enc in most cases, including the one you used, does password-based encryption.
This does not encrypt or decrypt with the "key" you specify, instead it considers that a password 
and runs it through a "stretching" or "derivation" function to produce the actual key. 
See https://stackoverflow.com/questions/22085107/perl-cbc-des-equivalent-in-java 
and https://stackoverflow.com/questions/11783062/how-to-decrypt-an-encrypted-file-in-java-with-openssl-with-aes .
For "direct" encryption and decryption you must use -K (uppercase) with the key in hex.

2: the example data in the RFC is in hex, although it doesn't say so.
For the key and IV this is okay because enc -K -iv take hex, but the data must be binary.
Hex 313233...30 is ASCII 12345678901234567890 but with no terminating newline 
which echo automatically adds; preventing this varies depending on your system and/or shell, which 
you don't specify, so I use perl instead. 

3: XMLenc data for block/CBC contains the (actual binary) IV concatenated with 
the ciphertext, with the combined result encoded in base64.

Thus (with binmode because I tested partly on Windows):

  echo $k
  12345678901234567890123456789012
  echo $i
  000102030405060708090a0b0c0d0e0f
  echo $i |\
  perl -ne "open O,'&gt;temp';binmode O;print O map{pack('H2',$_)} ($_=~/(..)/g)" 
  # that puts the binary IV at the beginning of temp
  perl -e "print '12345678901234567890'" |\
  openssl enc -aes-128-cbc -K $k -iv $i &gt;&gt;temp
  # that APPENDS the encryption of the data to temp
  openssl base64 &lt;temp
  AAECAwQFBgcICQoLDA0OD+cIHItlB3Wra1DUpxVvOx2lef1VmNPCMl8jwZqIUqGv


4: MAC is computed on the encrypted value; also dgst -hmac takes the key as actual bytes (not hex) 
in a C string, but the MAC key in this example contains a null byte which is impossible in a C string.
If you really want, you can do the HMAC as two hashes with tweaks of the key:

  echo $h |\
  perl -ne "open O,'&gt;pad1';binmode O;$_=~s/[[:space:]]//g;\
  $_.='0'x(128-length($_));print O map{pack 'C',hex($_)^0x36}($_=~/(..)/g)"
  echo $h |\
  perl -ne "open O,'&gt;pad2';binmode O;$_=~s/[[:space:]]//g;\
  $_.='0'x(128-length($_));print O map{pack 'C',hex($_)^0x5C}($_=~/(..)/g)"
  cat pad1 temp |openssl sha1 -binary -out tmp1
  cat pad2 tmp1 |openssl sha1 -binary -out tmp2
  openssl base64 &lt;out2
  Su+NvtQfmvfJzF6bmQiJqoLRExc=


And there you are.