Part 1. A chemist reacted 12.0 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 280. K and 1.50 atm. F2 + 2NaCl → Cl2 + 2NaF

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Part 1. A chemist reacted 12.0 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 280. K and 1.50 atm. F2 + 2NaCl → Cl2 + 2NaF

1. 91.728g of NaCl

2. 62.712g of NaCl

Explanation:

Part 1:

We obtained the following information from the question:

V = 12L

P = 1.5atm

T = 280K

R = 0.082atm.L/K /mol

n

PV = nRT

n = PV /RT

n = (12x1.5) / (0.082x280)

n = 0.784mol

Converting the number of mole of F2 obtained to mass, we have:

MM of F2 = 2 x 19 = 38g/mol

Mass of F2 = 0.784 x 38 = 29.792g

F2 + 2NaCl → Cl2 + 2NaF

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass conc of NaCl from the balanced equation = 2 x 58.5 = 117g

From the equation,

38g of F2 reacted with 117g of NaCl.

Therefore

29.792g of F2 will react with = (29.792 x 117) / 38 = 91.728g of NaCl

Part 2:

1mole of a gas occupy 22.4L at stp. This implies that 1mole of F2 also occupy 22.4L at stp. Now, we must determine how many moles of F2 will occupy 12L as given from the question. This can be achieved by doing the following:

22.4L = 1mol

12L = 12/22.4 = 0.536mol

Now let us convert this mole of F2 (i.e 0.536mol) to grams.

Mass of F2 = 0.536 x 38 = 20.368g

Now let us consider the equation.

From the equation,

38g of F2 reacted with 117g of NaCl.

Therefore,

20.368g of F2 will react with = (20.368x117)/38 = 62.712g of NaCl

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- [tex]m_{NaCl}=91.65gNaCl[/tex]

- STP: [tex]m_{NaCl}=62.63gNaCl[/tex]

Explanation:

Hello,

In this case, for the given reaction, we can notice a 1:2 mole ratio between fluorine and sodium chloride, thus, for the given volume, temperature and pressure, we compute the moles of reacting fluorine:

[tex]n_{F_2}=\frac{PV}{RT}=\frac{12.0L*1.50atm}{0.082\frac{atm*L}{mol*K}*280K}=0.784molF_2[/tex]

Therefore, given the aforementioned 1:2 mole ratio and the molar mass of sodium chloride (58.45 g/mol) we can compute its reacting mass:

[tex]m_{NaCl}=0.784molF_2*\frac{2molNaCl}{1molF_2} *\frac{58.45gNaCl}{1molNaCl} \\\\m_{NaCl}=91.65gNaCl[/tex]

Now, at STP, we must change both temperature and pressure to 273.15 K and 1.00 atm respectively, so the reacting moles of fluorine for the same volume are:

[tex]n_{F_2}=\frac{PV}{RT}=\frac{12.0L*1.00atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.536molF_2[/tex]

And the mass of sodium chloride via the same previous procedure:

[tex]m_{NaCl}=0.536molF_2*\frac{2molNaCl}{1molF_2} *\frac{58.45gNaCl}{1molNaCl} \\\\m_{NaCl}=62.63gNaCl[/tex]

Regards.

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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.

2. Answer;

78.32 g

Explanation;

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.