Part 1. a chemist reacted 15.0 liters of f2 gas with nacl in the laboratory to form cl2 and naf. use the ideal gas law equation to determine the mass of nacl that reacted with f2 at 280. k and 1.50 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.

asap

1- 114.54 g.

2- 78.32 g.

Explanation:

Q1:

F₂ + 2NaCl → Cl₂ + 2NaF.

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).

∴ n = PV/RT = (1.5 atm)(15.0 L) / (0.082 L/atm/mol.K)(280.0 K) = 0.98 mol.

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.

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.98 mole)/(1.0 mole) = 1.96 mol.

∴ The mass of NaCl = n x molar mass of NaCl = (1.96 mol)(58.44 g/mol) = 114.54 g.

Q2:

∴ n = PV/RT = (1.0 atm)(15.0 L) / (0.082 L/atm/mol.K)(273.0 K) = 0.67 mol.

Using cross multiplication:

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

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

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.67 mole)/(1.0 mole) = 1.34 mol.

∴ The mass of NaCl = n x molar mass of NaCl = (1.34 mol)(58.44 g/mol) = 78.32 g.

the answer for sodium is 0.186

the answer for rubidium is 0.251

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).