Answer:
Part 1
The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams
Part 2
The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams
Explanation:
Part 1
The volume of F₂ gas in the reaction, V = 18.0 liters
The ideal gas equation is P·V = n·R·T
∴ n = P·V/(R·T)
The pressure, P = 1.5 atm
The temperature, T = 290 K
The universal gas constant, R = 0.0820573 L·atm/(mol·K)
∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615
The number of moles of F₂ in the reaction n ≈ 1.134615 moles
The chemical reaction is given as follows;
F₂ + 2NaCl → Cl₂ + 2NaF
1 mole of F₂ reacts with 2 moles of NaCl
Therefore;
1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl
1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol
The mass, of 2.26923 moles of NaCl, m = Number of moles × MM
∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams
The mass of the NaCl ≈ 132.6 gams
Part 2
The volume occupied by 1 mole of all gases at STP = 22.4 l/mole
Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles
Therefore;
The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles
The number of moles of NaCl, in the reaction n ≈ 1.608 moles
The mass of NaCl in the reaction, m = n × MM
∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams
The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams
= 
((0.693 X 162.5 ) / 32.5) = 121.5