Question

A chemist reacted 30.0 liters of F2 gas with NaCl in the laboratory to form Cl2 and NaF. Use the ideal gas law equation to determine the mass of NaCl that reacted with F2 at 300 K and 1.50 atm. F2 + 2NaCl → Cl2 + 2NaF

Answer 1

The Ideal gas law is as follows:

`PV=nRT`

Since we have the informationof how many liters of F₂ we have as well as the temperature and pressure conditions, we can calculate how many moles of F₂ reacted:

`n=(PV)/(RT)`
`\begin{gathered} P=1.50atm \\ V=30.0L \\ T=300K \end{gathered}`

Since ew have the pressure in atm and the volume in L, we can use the following R constant:

`R\approx0.0820574L\cdot atm\cdot K^(-1)\cdot mol^(-1)`

So, we have:

`\begin{gathered} n=\frac{1.50atm\cdot30.0L}{0.0820574L\cdot atm\cdot K^(-1)_{}\cdot mol^(-1)\cdot300K} \\ n=(1.50\cdot30.0)/(0.0820574\cdot300)mol \\ n=(45)/(24.61722)mol \\ n=1.8279\ldots mol \end{gathered}`

Now, usin the balanced reaction:

`F_2+2NaCl\to Cl_2+2NaF`

We can see that for each mol of F₂ that reacts, 2 mols of NaCl will react, so:

`\begin{gathered} n_(F_2)=1.8279\ldots mol_{} \\ n_(NaCl)=2n_(F_2)=2\cdot1.8279\ldots mol_{}=3.6559\ldots mol \end{gathered}`

Now, to get this value in mass, we will need the molar mass of NaCl:

`M_(NaCl)=1\cdot M_(Na)+1\cdot M_(Cl)_{}=(1\cdot22.98976928+1\cdot35.453)g\/mol=58.44276928g\/mol`

Now, we can calculate the mass of NaCl:

`\begin{gathered} M_(NaCl)=(m_(NaCl))/(n_(NaCl)) \\ m_(NaCl)=n_(NaCl)\cdot M_(NaCl)=3.6559\ldots mol\cdot58.44276928g\/mol=213.66\ldots g\approx214g \end{gathered}`

So, approximately 214 g of NaCl wil react.