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

Approximately
`0.110\; \rm mol`.

Step-by-step 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 &= (m)/(M) \\ &\approx (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})} = (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 * (1)/(5) \approx 0.110\; \rm mol\end{aligned}`.