Chemistry, 31.01.2020 23:04, sjjarvis53211

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.

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answered: isaacbryan2416

Answer :

For part 1 : The mass of NaCl reacted is, 114.543 grams.

For part 1 : The mass of NaCl reacted is, 78.273 grams.

Explanation :

For part 1 :

First we have to calculate the volume of for mole of F_2 gas by using ideal gas equation.

PV=nRT

where,

P = pressure of F_2 gas = 1.50 atm

V = volume of F_2 gas = 15.0 L

T = temperature of F_2 gas = 280 K

n = number of moles of F_2 gas = ?

R = gas constant = 0.0821L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

$(1.50atm)\times (15.0L)=n\times (0.0821L.atm/mole.K)\times (280K)$

n=0.979mole

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

$F_2+2NaCl\rightarrow Cl_2+2NaF$

From the balanced reaction, we conclude that

As, 1 mole of F_2 gas react with 2 moles of NaCl

So, 0.979 mole of F_2 gas react with $2\times 0.979=1.958$ moles of NaCl

Now we have to calculate the mass of NaCl.

$\text{Mass of }NaCl=\text{Moles of }NaCl\times \text{Molar mass of }NaCl$

$\text{Mass of }NaCl=1.958mole\times 58.5g/mole=114.543g$

The mass of NaCl reacted is, 114.543 grams.

For part 2 :

First we have to calculate the volume of for mole of F_2 gas by using ideal gas equation.

PV=nRT

where,

P = pressure of F_2 gas = 1 atm (at STP)

V = volume of F_2 gas = 15.0 L

T = temperature of F_2 gas = 273 K (at STP)

n = number of moles of F_2 gas = ?

R = gas constant = 0.0821L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (15.0L)=n\times (0.0821L.atm/mole.K)\times (273K)$

n=0.669mole

Now we have to calculate the moles of NaCl.

The balanced chemical reaction is,

$F_2+2NaCl\rightarrow Cl_2+2NaF$

From the balanced reaction, we conclude that

As, 1 mole of F_2 gas react with 2 moles of NaCl

So, 0.669 mole of F_2 gas react with $2\times 0.669=1.338$ moles of NaCl

Now we have to calculate the mass of NaCl.

$\text{Mass of }NaCl=\text{Moles of }NaCl\times \text{Molar mass of }NaCl$

$\text{Mass of }NaCl=1.338mole\times 58.5g/mole=78.273g$

The mass of NaCl reacted is, 78.273 grams.

answered: Guest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.