How are the ions in ionic compounds arranged?​

1.

114.54 g.

Explanation:

The balanced equation of the reaction is:

F₂ + 2NaCl → Cl₂ + 2NaF.

From the equation;

1.0 mole of F₂ reacts with 2.0 moles of NaCl to produce 1.0 mole of Cl₂ and 2.0 moles of NaF.

The number of moles of F₂ (15.0 liters at 280.0 K and 1.50 atm) can be calculated using the ideal gas law:

PV = nRT; where, P is the pressure of the gas (P = 1.0 atm), V is the volume of the gas (V = 15.0 L), n is the no. of moles of the gas (n = ??? mole), R is the general gas constant (R = 0.082 L.atm/mol.K), T is the temperature of the gas (T = 280.0 K).

Therefore;

n = PV/RT

   = (1.5 atm)(15.0 L) / (0.082 L/atm/mol.K)(280.0 K)

    = 0.98 mol.

The number of moles of NaCl  that reacted with 0.98 mol of F₂, wiil be

Using cross multiplication;

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.98 mole of F₂ reacts with → ??? moles of NaCl.

we get;

       = (2.0 mole)(0.98 mole)/(1.0 mole)

       = 1.96 mol.

Hence; the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

                                 = n x molar mass of NaCl

                                = (1.96 mol)(58.44 g/mol)

                                = 114.54 g.

78.32 g

STP conditions means that P = 1.0 atm and T = 273.0 K.

Calculating the no. of moles of F₂ (15.0 liters at 273.0 K and 1.0 atm) using the ideal gas law: PV = nRT,

Thus;

n = PV/RT

  = (1.0 atm)(15.0 L) / (0.082 L/atm/mol.K)(273.0 K)

  = 0.67 moles

We can get the no. of moles of NaCl  reacted with 0.67 mol of F₂.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stoichiometry.

0.67 mole of F₂ reacts with → ??? moles of NaCl.

Thus;

The no, of moles of reacted NaCl;

= (2.0 mole)(0.67 mole)/(1.0 mole)

= 1.34 mol.

Then, we can get the mass of NaCl that reacted with F₂ at 280.0 K and 1.50 atm:

   = n x molar mass of NaCl

   = (1.34 mol)(58.44 g/mol)

   = 78.32 g.