Answer:
141g of CCl₄
Explanation:
First, we have to write the balanced equation.
CCl₄(g) + 2 HF(g) ⇄ CF₂Cl₂(g) + 2 HCl(g)
We can calculate how many moles of CF₂Cl₂ using the ideal gas equation.
V = 14.9 dm³ = 14.9 L
T = 21°C + 273.15 = 294.15 K
P = 1.48 atm
R = 0.08206 atm.L/mol.K
[tex]P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.48atm.14.9L}{0.08206\frac{atm.L}{mol.K}.294.15K }=0.914mol[/tex]
We can use proportions to find the mass of CCl₄ required to obtain 0.914 moles of CF₂Cl₂. According to the balanced equation, 1 mol of CF₂Cl₂ is produced when 1 mol of CCl₄ reacts. And the molar mass of CCl₄ is 154 g/mol.
[tex]0.914molCF_{2}Cl_{2}.\frac{1molCCl_{4}}{1molCF_{2}Cl_{2}} .\frac{154gCCl_{4}}{1molCCl_{4}} =141gCCl_{4}[/tex]
