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jonny [76]
3 years ago
8

In a reaction, 24.9 L of N2 reacts with excess Hy to produce NH3. How many liters of NH3

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

A. 49.8L of NH3.

B. 33.83g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

N2 + 3H2 —> 2NH3

A. From the balanced equation above,

1L of N2 produced 2L of NH3.

Therefore, 24.9L of N2 will produce = 24.9 x 2 = 49.8L of NH3.

Therefore, 49.8L of NH3 is produced from the reaction.

B. Determination of the mass NH3 produced.

First, we shall determine the number of mole of NH3 produced. This can be obtained as follow:

Volume (V) = 49.8L

Pressure (P) = 97.8 kPa = 97.8/101.325 = 0.97atm

Temperature (T) = 23.7°C = 23.7°C + 273 = 296.7K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) =...?

PV = nRT

n = PV /RT

n = (0.97 x 49.8) / (0.082 x 296.3)

n = 1.99 mole

Next, we shall convert 1.99 mole to NH3 to grams. This is illustrated below:

Number of mole NH3 = 1.99 mole

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 1.99 x 17

Mass of NH3 = 33.83g

Therefore, 33.83g of NH3 is produced.

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The solubility of carbon dioxide at 400 kPa  at room temperature is ;

( B ) 0.61 CO2/L

<u>Given data </u>

pressure of CO₂ = 400 Kpa = 3.95 atm

Kh of CO₂ = 3.3 * 10⁻² mol/L.atm

<h3>Calculate the solubility of carbon dioxide </h3>

Solubility = pressure * Kh value of CO₂

                = 3.95 atm * 3.3 * 10⁻² mol / L.atm

                = 0.13 mol/l  CO₂

                = 0.61 CO₂ / L

Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l  CO₂.

Learn more about solubility : brainly.com/question/23946616

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2.00 moles of an ideal gas was found to occupy a volume of 17.4L at a pressure of 3.00 atm and at a temperature of 45 C. Calulat
lidiya [134]

Answer:

0.082 atm L mol^{-1} K^{-1}

Explanation:

The pressure, the volume and the temperature of an ideal gas are related to each other by the equation of state:

pV=nRT

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles

R is the gas constant

T is the absolute temperature

For the gas in this problem:

n = 2.00 mol is the number of moles

V = 17.4 L is the gas volume

p = 3.00 atm is the gas pressure

T=45C+273=318 K is the absolute temperature

Solving for R, we find the gas constant:

R=\frac{pV}{nT}=\frac{(3.00)(17.4)}{(2.00)(318)}=0.082 atm L mol^{-1} K^{-1}

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