What are the products for the redox reaction HCl + SnCl₂ + KMnO₄?
When using $\ce{MnO4-}$ as an oxidant, pH is crucial.

In acidic medium: $\ce{Mn(VII) -&gt; Mn(II)}$

$$\ce{8 H3O+ + MnO4^{-} + 5 e- -&gt; Mn^{2+} + 12 H2O}$$

Under neutral conditions: $\ce{Mn(VII) -&gt; Mn(IV)}$

$$\ce{2 H2O + MnO4^{−} + 3 e− -&gt; MnO2 + 4 OH^{−}}$$

In strongly alkaline medium: $\ce{Mn(VII) -&gt; Mn(VI)}$

$$\ce{MnO4^{−} + e− -&gt; MnO4^{2−}}$$

As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes 
$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.