Question

PC15 (8) + H₂O(1) → POCl3 (1) + 2HCl(aq)When 64.23 g of phosphorus pentachloride reacts with ewhat mass of hydrogen chloride will be produced?If needed, enter scientific notation with the "e". For examwould be entered as 1.44e7.Answer:g HCI.

Answer 1

The reaction of 64.23 g of
`\(PCl_5\)` with excess water produces 22.51 g of
`\(HCl\)`

In scientific notation:
`\(2.251 * 10^1 \, \text` g HCl

To determine the mass of hydrogen chloride produced when 64.23 g of phosphorus pentachloride
`(\(PCl_5\))` reacts with excess water, we can use stoichiometry and the balanced chemical equation:

`\[ PCl_5 (s) + H_2O(l) \rightarrow POCl_3 (l) + 2HCl(aq) \]`

First, find the molar mass of
`\(PCl_5\)` and
`\(HCl\)`:

`\text{Molar mass of } PCl_5 &= 30.97 (\text{P}) + 5 * 35.45 (\text{Cl}) = 208.22 \, \text{g/mol}`

`\text{Molar mass of } HCl &= 1.01 (\text{H}) + 35.45 (\text{Cl}) = 36.46 \, \text{g/mol}`

Next, determine the moles of
`\(PCl_5\)` using its given mass:

`\[\text{Moles of } PCl_5 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{64.23 \, \text{g}}{208.22 \, \text{g/mol}} = 0.3087 \, \text{mol}\]`

According to the balanced equation, 1 mole of
`(\(PCl_5\))` produces 2 moles of
`\(HCl\)`. Therefore, moles of
`\(HCl\)` produced:

`\[\text{Moles of } HCl = 2 * \text{Moles of } PCl_5 = 2 * 0.3087 \, \text{mol} = 0.6174 \, \text{mol}\]`

Finally, convert moles of
`\(HCl\)` to grams:

`\[\text{Mass of } HCl = \text{Moles} * \text{Molar mass} = 0.6174 \, \text{mol} * 36.46 \, \text{g/mol} = 22.51 \, \text{g}\]`

So, when 64.23 g of
`\(PCl_5\)` reacts with excess water, 22.51 g of
`\(HCl\)` will be produced.

Answer 2

22.52g of hydrogen chloride will be produced.

Step-by-step explanation:

1st) From the balanced reaction we know that 1 mole of PCl5 reacts with 1 mole of water to produce 1 mole of POCl3 and two moles of hydrogen chloride. Using the molar mass of PCl5 (208.24g/mol) and HCl (36.5g/mol) we can convert the moles to grams:

- 1 mole of PCl5 weighs 208.24g.

- 2 moles of HCl weigh 73g

2nd) With the grams of PCl5 and HCl from the stoichiometry, and the given value of 64.23g of PCl5, we can calculate the mass of HCl that will be produced:

`\begin{gathered} 208.24gPCl_5-73gHCl \\ 64.23gPCl_5-x=(64.23gPCl_5*73gHCl)/(208.24gPCl_5) \\ x=22.52gHCl \end{gathered}`

Finally, 22.52g of hydrogen chloride will be produced.