Answer:
28.22L of HF are needed.
Explanation:
1st) It is necessary to convert the 14.2L H2 to moles, using the relation for gases that 22.4L is equal to the volume of 1 mole:
[tex]\begin{gathered} 22.4L-1mol \\ 14.2L-x=\frac{14.2L*1mol}{22.4L} \\ x=0.63moles \end{gathered}[/tex]2nd) Now, with the stoichiometry of the balanced reaction (we know that 1 mole of H2 is formed from 2 moles of HF), and the 0.63 moles of H2, we can calculate the moles of HF needed:
[tex]\begin{gathered} 1molH_2-2molHF \\ 0.63molH_2-x=\frac{0.63molH_2*2molHF}{1molH_2} \\ x=1.26molHF \end{gathered}[/tex]Now we know that 1.26 moles of HF can produce 14.2L (0.63moles) of H2 gas.
3rd) Finally, we have to convert the 1.26 moles of HF to liters:
[tex]\begin{gathered} 1mole-22.4L \\ 1.26moles-x=\frac{1.26moles*22.4L}{1mole} \\ x=28.22L \end{gathered}[/tex]So, 28.22L of HF are needed.