The picture below shows two bulbs connected by a stopcock. the large bulb, with a volume of 6.00 l, contains nitric oxide at a pressure of 0.500 atm, and the small bulb, with a volume of 1.50 l, contains oxygen at a pressure of 2.50 atm. the temperature at the beginning and the end of the experiment is 22 °c. what are the partial gasses of no, and no2?

Answers:  

p(NO) = 0; p(O₂) = 0.300 atm; p(NO₂) = 0.400 atm  

Step-by-step explanation:  

Step 1. Moles of NO  

pV = nRT  

n = (pV)/(RT)  

p = 0.500 atm  

R = 0.082 06 L·atm·K⁻¹mol⁻¹  

T = (22 + 273.15)  

T = 295.15 K  

n = (0.500 × 6.00)/(0.082 06 × 295.15)  

n = 3.00/24.22  

n = 0.1239 mol  

Step 2. Moles of O₂  

p = 2.50 atm  

V = 1.50 L  

n = (2.50 × 1.50)/24.22  

n = 3.75/24.22  

n = 0.1548 mol  

Step 3. Identify the limiting reactant  

             2NO  +   O₂  ⟶  2NO₂  

n/mol: 0.1239  0.1548  

Calculate the moles of NO₂ we can obtain from each reactant.    

From NO:  

The molar ratio of NO₂: NO is 2:2

Moles of NO₂ = 0.1239 × 2/2  

Moles of NO₂ = 0.1239 mol NO₂  

From O₂:  

The molar ratio of NO₂: O₂ is 2:1.  

Moles of NO₂ = 0.1548 × 2/1  

Moles of NO₂ = 0.3097 mol NO₂  

NO is the limiting reactant because it gives the smaller amount of NO₂.  

Step 4. Moles of each species after reaction  

               2NO  +       O₂  ⟶  2NO₂  

I/mol:     0.1239      0.1548         0  

C:/mol: -0.1239     -0.06193  +0.1239  

F/mol:        0          0.0929      0.1239  

Step 5. Partial pressure of NO  

p(NO) = 0  

Step 6. Partial pressure of O₂  

pV = nRT  

p = (nRT)/V  

V= 6.00 + 1.50  

V = 7.50 L  

p = (0.0929 × 24.22)/7.50  

p = 2.25/7.50  

p = 0.300 atm

Step 7. Partial pressure of O₂  

p = (0.1239 × 24.22)/7.50  

p = 3.00/7.50  

p = 0.400 atm  

p(NO) = 0; p(O₂) = 0.300 atm; p(NO₂) = 0.400 atm  

The partial pressure of NO is 0 atm, and is 0.400 atm at the end of the reaction.

The concentration of NO and initially are;

By ideal gas equation:

PV =nRT

P = pressure

V = volume

n = moles

R = constant = 0.0816 L.atm/mol.K

T = temperature = 22 = 273 + 22 K = 295 K

Moles of NO =

Moles of  NO =

Moles of NO = 0.1239 moles

Moles of =

Moles of = 0.1548 moles

The reaction will be:

2 moles of NO gives 2 moles of

So, 0.1239 moles of NO gives 0.1239 moles of

1 mole of gives 2 moles of

0.1548 moles of gives 0.3097 moles of

Since No produces smaller amount of , it is the limiting factor.

Partial pressure of NO = 0 since it is completely used.

Partial pressure of =

n = 0.1239 moles

Total volume = 6 + 1.50

V = 7.5 L

Partial pressure of =

Partial pressure of = 0.400 atm

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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 and electronic configuration of phosphorous is .

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.

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.

the answer for sodium is 0.186

the answer for rubidium is 0.251