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2 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a reactor. What is the excess reactant?

Chemistry

1 answer:

Grace [21]2 years ago

3 0

Answer:

H₂ gas

Explanation:

The reaction between nitrogen gas and hydrogen gas forms ammonia (the Haber-Bosch process):

N₂ + 3H₂ ⇒ 2NH₃

The excess reactant can be found by comparing the moles of nitrogen and hydrogen. The molar mass of N₂ is 28.00 g/mol and the molar mass of H₂ is 2.02 g/mol.

(100 kg N₂)(1000g/kg)(mol/28.00g) = 3570 mol

(100 kg H₂)(1000g/kg)(mol/2.02g) = 49500 mol

The molar ratio between the reactant N₂ and H₂ is 1N₂:3H₂. The moles of nitrogen required to react with H₂ is:

(49500 mol H₂)(1N₂ / 3H₂) = 16500 mol

The amount of nitrogen required is more than what is available, so nitrogen is the limiting reagent and hydrogen is the excess reagent.

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Any reaction that absorbs 150 kcal of energy can be classified as.

N76 [4]

Answer:

The Answer should be Endothermic reaction

Explanation:

Reactions in which energy is absorbed are called Endothermic reactions.

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2 years ago

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11111nata11111 [884]

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3 years ago

A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc

elena55 [62]

For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

%He=12.5%

%CH4=87.5%

<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

2.0/100*100=50/100*20+%*80

Hence

%He=12.5%

%CH4=87.5%

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3 0

1 year ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

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2 years ago

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2 years ago