Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equation to determine the mass of NaCl that reacted with F2 at 290. K and 1.5 atm. F2 + 2NaCl → Cl2 + 2NaF

Part 2. Explain how you would determine the mass of sodium chloride that can react with the same volume of fluorine gas at STP.

a) 132 g NaCl consumed at 1.5 Atm & 290K

b) 93.2 g NaCl consumed at STP

Explanation:

F₂ + 2NaCl => 2NaF + Cl₂

Given 18L F₂(g) at 1.5Atm & 290K

moles F₂(g) = PV/RT = (1.5Atm)(18L)/(0.08206L·Atm/mol·K)(290K) = 1.135 mol F₂(g) consumed

moles NaCl consumed = 2(moles F₂) = 2(1.135) mol NaCl = 2.269 mol NaCl

mass NaCl = 2.269 mol NaCl x 58 g/mol = 131.61 g ≅ 132 g consumed at 1.5 Atm & 290K

Given 18L F₂(g) - same volume as in part a -  at STP

moles F₂(g) = PV/RT = (1.0Atm)(18L)/(0.08206L·Atm/mol·K)(273K) = 0.803 mol F₂(g) consumed

moles NaCl consumed = 2(moles F₂) = 2(0.803) mole NaCl = 1.607 mol NaCl

mass NaCl = 1.607 mol NaCl x 58 g/mol = 93.2 g NaCl consumed at STP