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Without $\ce{NH3}$, $\ce{SO2}$ produces no effet on a $\ce{BaCl2}$ solution, because $\ce{SO2}$ produces an acidic solution where $\ce{BaSO3}$ is not formed. $\ce{NH3}$ prevents the solution to become acidic. In the presence of $\ce{NH3}$, a white $\ce{BaSO3}$ precipitate is produced according to : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}\tag{1}$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + 2 NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

You think that $\ce{BaSO4}$ may be formed. It will not happen without air. In the presence of air, both $\ce{BaSO3}$ and $\ce{SO2}$ may be oxidized into $\ce{BaSO4}$ and $\ce{H2SO4}$ according to : $$\ce{2 BaSO3 + O2 -> 2 BaSO4}$$ $$\ce{2SO2 + O2 + 2 H2O -> 2 H2SO4}$$ and of course $\ce{H2SO4}$ will produce a $\ce{BaSO4}$ precipitate in the $\ce{BaCl2}$ solution.

Without $\ce{NH3}$, $\ce{SO2}$ produces no effect on a $\ce{BaCl2}$ solution, because $\ce{SO2}$ produces an acidic solution where $\ce{BaSO3}$ is not formed. $\ce{NH3}$ prevents the solution becoming acidic. In the presence of $\ce{NH3}$, a white $\ce{BaSO3}$ precipitate is produced according to : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}\tag{1}$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + 2 NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

You think that $\ce{BaSO4}$ may be formed. It will not happen without air. In the presence of air, both $\ce{BaSO3}$ and $\ce{SO2}$ may be oxidized into $\ce{BaSO4}$ and $\ce{H2SO4}$ according to : $$\ce{2 BaSO3 + O2 -> 2 BaSO4}$$ $$\ce{2SO2 + O2 + 2 H2O -> 2 H2SO4}$$ and of course $\ce{H2SO4}$ will produce a $\ce{BaSO4}$ precipitate in the $\ce{BaCl2}$ solution.

Without $\ce{NH3}$, $\ce{SO2}$ produces no effet on a $\ce{BaCl2}$ solution, because $\ce{SO2}$ produces an acidic solution where $\ce{BaSO3}$ is not formed. $\ce{NH3}$ prevents the solution to become acidic. In the presence of $\ce{NH3}$, a white $\ce{BaSO3}$ precipitate is produced according to : : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}..........(\tag(1)$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + 2 NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

You think that $\ce{BaSO4}$ may be formed. It will not happen without air. In the presence of air, both $\ce{BaSO3}$ and $\ce{SO2}$ may be oxidized into $\ce{BaSO4}$ and $\ce{H2SO4}$ according to : $$\ce{2 BaSO3 + O2 -> 2 BaSO4}$$ $$\ce{2SO2 + O2 + 2 H2O -> 2 H2SO4}$$ and of course $\ce{H2SO4}$ will produce a $\ce{BaSO4}$ precipitate in the $\ce{BaCl2}$ solution.

Without $\ce{NH3}$, $\ce{SO2}$ produces no effet on a $\ce{BaCl2}$ solution, because $\ce{SO2}$ produces an acidic solution where $\ce{BaSO3}$ is not formed. $\ce{NH3}$ prevents the solution to become acidic. In the presence of $\ce{NH3}$, a white $\ce{BaSO3}$ precipitate is produced according to : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}\tag{1}$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + 2 NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

You think that $\ce{BaSO4}$ may be formed. It will not happen without air. In the presence of air, both $\ce{BaSO3}$ and $\ce{SO2}$ may be oxidized into $\ce{BaSO4}$ and $\ce{H2SO4}$ according to : $$\ce{2 BaSO3 + O2 -> 2 BaSO4}$$ $$\ce{2SO2 + O2 + 2 H2O -> 2 H2SO4}$$ and of course $\ce{H2SO4}$ will produce a $\ce{BaSO4}$ precipitate in the $\ce{BaCl2}$ solution.

$\ce{NH3}$ prevents the solution to become acidic. The only effect of $\ce{SO2}$ in presence of $\ce{NH3}$ is the production of a white $\ce{BaSO3}$ precipitate according to : : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}..........(\tag(1)$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

Without $\ce{NH3}$, $\ce{SO2}$ produces no effet on a $\ce{BaCl2}$ solution, because $\ce{SO2}$ produces an acidic solution where $\ce{BaSO3}$ is not formed. $\ce{NH3}$ prevents the solution to become acidic. In the presence of $\ce{NH3}$, a white $\ce{BaSO3}$ precipitate is produced according to : : $$\ce{Ba^{2+} + SO2 + H2O -> BaSO3 + 2 H+}..........(\tag(1)$$ followed here by $$\ce{NH3 + H+ -> NH4+}$$ so that the total reaction is more correctly written : $$\ce{Ba^{2+} + SO2 + 2 NH3 + H2O -> BaSO3 + 2 NH4+ }$$ If $\ce{NH3}$ was missing in this mixture, the reaction ($1$) would not take place. It would even be reversed, and, instead of being formed, $\ce{BaSO3}$ would be dissolved by $\ce{H+}$ producing $\ce{Ba^{2+}, SO2}$ and $\ce{H2O}$.

You think that $\ce{BaSO4}$ may be formed. It will not happen without air. In the presence of air, both $\ce{BaSO3}$ and $\ce{SO2}$ may be oxidized into $\ce{BaSO4}$ and $\ce{H2SO4}$ according to : $$\ce{2 BaSO3 + O2 -> 2 BaSO4}$$ $$\ce{2SO2 + O2 + 2 H2O -> 2 H2SO4}$$ and of course $\ce{H2SO4}$ will produce a $\ce{BaSO4}$ precipitate in the $\ce{BaCl2}$ solution.