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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