= Solution
There is no infinite component in the modulus $(5)$. The ordinary <ideal class group> of $\mathbb Q$ is trivial, and its <unit group> is $\{\pm1\}$. The <ray class exact sequence> therefore gives
$$
\boxed{\operatorname{Cl}_{(5)}(\mathbb Q)
\simeq(\mathbb Z/5\mathbb Z)^\times/\{\pm1\}\simeq C_2.}
$$
Concretely, a <fractional ideal> prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.
To identify the corresponding field, put $E=\mathbb Q(\zeta_5+\zeta_5^{-1})=\mathbb Q(\sqrt5)$. The <maximal real subfield of the fifth cyclotomic field> is fixed by complex conjugation, the element $-1$ of $(\mathbb Z/5\mathbb Z)^\times$. For a prime $\ell\ne5$, the <Frobenius automorphism> of $\mathbb Q(\zeta_5)/\mathbb Q$ is $\zeta_5\mapsto\zeta_5^\ell$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm1\}$. Multiplicativity extends this to the <Artin reciprocity map> on all <fractional ideals> prime to $5$, whose kernel is exactly $P_{\mathbb Q,1}((5))$. Its <field discriminant> $5$ also confirms that $E$ is <unramified> outside $5$; moreover $E$ is a <totally real number field>. The uniqueness in the <Main theorem of class field theory> now gives
$$
\boxed{\mathbb Q_{(5)}=\mathbb Q(\sqrt5).}
$$
Including the real place changes the <ray class group>: a principal generator must then be positive as well as congruent to one. The <residue and signature group of a modulus> gives
$$
\operatorname{Cl}_{(5)\infty}(\mathbb Q)
\simeq\bigl((\mathbb Z/5\mathbb Z)^\times\times\{\pm1\}\bigr)
/\langle(-1,-1)\rangle
\simeq(\mathbb Z/5\mathbb Z)^\times.
$$
Here the isomorphism sends $(u,\varepsilon)$ to $\varepsilon u$. Consequently $\mathbb Q_{(5)\infty}=\mathbb Q(\zeta_5)$, of degree four. \b[The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field.] The printed finite-modulus assertion needs no correction.
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