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boyakko [2]
2 years ago
7

A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is

a stainless-steel cylinder that measures wide 22cm and 26.4 high. The maximum safe pressure inside the vessel has been measured to be 6.30mpa.
For a certain reaction, the vessel may contain up to .537kg of dinitrogen monoxide gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
5 0

The maximum safe operating temperature for this reaction is equal to 895°C.

<u>Given the following data:</u>

  • Width of cylinder = 22 cm.
  • Height, h = 26.4 cm.
  • Maximum safe pressure = 6.30mpa.
  • Mass = 0.537 kg.

<u>Scientific data:</u>

  • Ideal gas constant, R = 8.314 L-kPa/Kmol.
  • Molar mass of of dinitrogen monoxide (N_2F_2) gas = 66 g/mol.

Radius, r = \frac{width}{2} =\frac{22}{2} =11\;cm

<h3>How to calculate the maximum safe operating temperature.</h3>

First of all, we would determine the volume of the stainless-steel cylinder by using this formula:

V=\pi r^2 h\\\\V = 3.142 \times 11^2 \times 26.4\\\\

Volume, V = 10,036.81 cm^3.

In liters, we have:

Volume, V = 10.04 Liters.

Next, we would determine the number of moles of dinitrogen monoxide (N_2F_2) gas:

Number \;of \;moles = \frac {mass}{molar\;mass}\\\\Number \;of \;moles = \frac {537}{66}

Number of moles = 8.136 moles.

Now, we can solve for the maximum safe operating temperature by applying the ideal gas equation:

PV=nRT\\\\T=\frac{PV}{nR} \\\\T=\frac{6.30 \times 10^3 \times 10.04}{8.136 \times 8.314}\\\\T=\frac{60541.2}{67.6427}

T = 895.02 ≈ 895°C.

Read more on temperature here: brainly.com/question/24769208

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Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how
Afina-wow [57]

Answer:

11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

Explanation:

1. The balanced chemical equation is the following:

Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)

2. Use the molar mass of the H_{2}O, the molar mass of the NH_{3} and the stoichiometry of the balanced chemical reaction to find how many grams of NH_{3} are produced:

Molar mass H_{2}O = 18\frac{g}{mol}

Molar mass NH_{3} = 17\frac{g}{mol}

36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}

Therefore 11.3 g of NH_{3} are produced from 36.0 g of H_{2}O

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