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ella [17]
3 years ago
15

What is the volume of 21.0 g of nitrogen gas at stp?

Chemistry
1 answer:
allochka39001 [22]3 years ago
7 0
We can use the ideal gas law equation to find the volume of N₂ gas at STP
PV = nRT
where 
P - standard pressure - 101 325 Pa
V - volume 
n - number of moles - 21.0 g / 28 g/mol = 0.750 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - standard temperature - 273 K
substituting the values in the equation 
101 325 Pa x V = 0.750 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 16.8 L
volume of N₂ at STP is 16.8 L
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vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

  • P1= 16 atm
  • V1
  • P2= ?
  • V2= V1÷2= \frac{V1}{2} because the volume is halved.

So:

16 atm*V1= P2* \frac{V1}{2}

Solving:

\frac{16 atm*V1*2}{V1}=P2

16 atm*2= P2

32 atm= P2

<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>

2. Now

  • P1= 5.5 atm
  • V1
  • P2= ?
  • V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

\frac{5.5 atm*V1}{3*V1}=P2

\frac{5.5 atm}{3}= P2

1.83 atm= P2

<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>

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What information do Quantum models show?
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Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:
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Answer:  1) Maximum mass of ammonia  198.57g  

2) The element that would be completely consumed is the N2

3) Mass that would keep unremained, is the one of  the excess Reactant, that means the H2 with 3,44g

Explanation:

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N2(g) + 3H2(g) ⟶2NH3(g)

Both equal amount of atoms side to side.

  • Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the  molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)

Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3

Moles of ammonia produced with 38.77 g H2⟶  38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3

  • As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element  that would be completey consumed, and the maximum mass of ammonia will be produced from it.
  • We proceed calculating the maximum mass of NH3 from the 163.3g of N2.

11.66  mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3

  • In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:

163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2

That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of  H2, thus, 38.77g - 35.33 g = 3.44g H2 is left

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