Question

How many moles of H 3 PO 4 are produced when 20.0 g of HCI are produced by the reaction PCl 5 +4H 2 O H 3 PO 4 +5HCl ?

Answer 1

Answer:

Approximately $0.110\; \rm mol$.

Explanation:

Balanced equation for this reaction:

$\rm PCl_5 + 4\; H_2O \to H_3PO_4 + 5\, HCl$.

Look up the relative atomic mass of $\rm H$ and $\rm Cl$ on a modern periodic table:

• $\rm H$: $1.008$.
• $\rm Cl$: $35.45$.

Calculate the molar mass of $\rm HCl$:

$\begin{aligned}&M({\rm HCl}) \\ &= (1.008 + 35.45)\; \rm g \cdot mol^{-1} \\ &\approx 36.46\; \rm g \cdot mol^{-1}\end{aligned}$.

Calculate the number of moles of molecules in that $20.0\; \rm g$ of $\rm HCl$:

$\begin{aligned}n &= \frac{m}{M} \\ &\approx \frac{20.0\; \rm g}{36.46\; \rm g \cdot mol^{-1}} \approx 0.549\; \rm mol\end{aligned}$.

In the balanced reaction, the coefficient of $\rm H_3PO_4$ and $\rm HCl$ are $1$ and $5$, respectively. The ratio between these two coefficients is:

$\displaystyle \frac{n({\rm H_3PO_4})}{n({\rm HCl})} = \frac{1}{5}$.

In other words, this reaction would produce five times as many $\rm HCl$ molecules as $\rm H_3PO_4$ molecules.

Calculation shows that in this question, approximately $0.549\; \rm mol$ of $\rm HCl$ molecules were produced. Calculate the number of moles of $\rm H_3PO_4$ molecules that were produced:

$\begin{aligned}& n({\rm H_3PO_4}) \\ &= n({\rm HCl}) \cdot \frac{n({\rm H_3PO_4})}{n({\rm HCl})} \\ & \approx 0.549\; \rm mol \times \frac{1}{5} \approx 0.110\; \rm mol\end{aligned}$.