Answer:
Approximately [tex]0.110\; \rm mol[/tex].
Explanation:
Balanced equation for this reaction:
[tex]\rm PCl_5 + 4\; H_2O \to H_3PO_4 + 5\, HCl[/tex].
Look up the relative atomic mass of [tex]\rm H[/tex] and [tex]\rm Cl[/tex] on a modern periodic table:
Calculate the molar mass of [tex]\rm HCl[/tex]:
[tex]\begin{aligned}&M({\rm HCl}) \\ &= (1.008 + 35.45)\; \rm g \cdot mol^{-1} \\ &\approx 36.46\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of moles of molecules in that [tex]20.0\; \rm g[/tex] of [tex]\rm HCl[/tex]:
[tex]\begin{aligned}n &= \frac{m}{M} \\ &\approx \frac{20.0\; \rm g}{36.46\; \rm g \cdot mol^{-1}} \approx 0.549\; \rm mol\end{aligned}[/tex].
In the balanced reaction, the coefficient of [tex]\rm H_3PO_4[/tex] and [tex]\rm HCl[/tex] are [tex]1[/tex] and [tex]5[/tex], respectively. The ratio between these two coefficients is:
[tex]\displaystyle \frac{n({\rm H_3PO_4})}{n({\rm HCl})} = \frac{1}{5}[/tex].
In other words, this reaction would produce five times as many [tex]\rm HCl[/tex] molecules as [tex]\rm H_3PO_4[/tex] molecules.
Calculation shows that in this question, approximately [tex]0.549\; \rm mol[/tex] of [tex]\rm HCl[/tex] molecules were produced. Calculate the number of moles of [tex]\rm H_3PO_4[/tex] molecules that were produced:
[tex]\begin{aligned}& n({\rm H_3PO_4}) \\ &= n({\rm HCl}) \cdot \frac{n({\rm H_3PO_4})}{n({\rm HCl})} \\ & \approx 0.549\; \rm mol \times \frac{1}{5} \approx 0.110\; \rm mol\end{aligned}[/tex].
