Respuesta :
The Ideal gas law is as follows:
[tex]PV=nRT[/tex]Since we have the informationof how many liters of Fâ we have as well as the temperature and pressure conditions, we can calculate how many moles of Fâ reacted:
[tex]n=\frac{PV}{RT}[/tex][tex]\begin{gathered} P=1.50atm \\ V=30.0L \\ T=300K \end{gathered}[/tex]Since ew have the pressure in atm and the volume in L, we can use the following R constant:
[tex]R\approx0.0820574L\cdot atm\cdot K^{-1}\cdot mol^{-1}[/tex]So, we have:
[tex]\begin{gathered} n=\frac{1.50atm\cdot30.0L}{0.0820574L\cdot atm\cdot K^{-1}_{}\cdot mol^{-1}\cdot300K} \\ n=\frac{1.50\cdot30.0}{0.0820574\cdot300}mol \\ n=\frac{45}{24.61722}mol \\ n=1.8279\ldots mol \end{gathered}[/tex]Now, usin the balanced reaction:
[tex]F_2+2NaCl\to Cl_2+2NaF[/tex]We can see that for each mol of Fâ that reacts, 2 mols of NaCl will react, so:
[tex]\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}[/tex]Now, to get this value in mass, we will need the molar mass of NaCl:
[tex]M_{NaCl}=1\cdot M_{Na}+1\cdot M_{Cl}_{}=(1\cdot22.98976928+1\cdot35.453)g\/mol=58.44276928g\/mol[/tex]Now, we can calculate the mass of NaCl:
[tex]\begin{gathered} M_{NaCl}=\frac{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}[/tex]So, approximately 214 g of NaCl wil react.
