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Arisa [49]
3 years ago
7

A liquid-fueled rocket engine has two liquids, usually called the fuel and the oxidizer. Which two liquids are most commonly use

d for this propellant?
Chemistry
1 answer:
hammer [34]3 years ago
8 0

Answer:

A liquid-fueled rocket has two liquids (liquids are good because of the density, they need less space than a gas to be stored), such that these liquids are called the fuel and the oxidizer.

These liquids are injected into a system that leads to a combustion chamber, where the liquids are mixed (we need to mix the fuel with the oxidizer to enable the combustion of the fuel) and burned to produce thrust.

Some common examples of oxidizers are liquid oxygen, which may be combined with fuels like liquid hydrogen, liquid methane, kerosene and hydrazine.

Other oxidizers are liquid fluorine (which also can be combined with the fuels liquid hydrogen and hydrazine), nitrogen tetroxide (which can be combined whit kerosene, hydrazine and other fuels) and FLOX-70, which can only be combined with kerosene.

The "most commonly used" may depend on the country and the type of liquid propellant ( petroleum, cryogens, and hypergols)

Such that the most common oxidizer may be liquid oxygen, and the most common fuel the kerosene.

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(2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) ) If 165 mL of oxygen is produced at 30.0 °C and 90.0 kPa, what mass of KClO3 was decomposed
soldier1979 [14.2K]

Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 KClO₃  → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • KClO₃: 2 moles  
  • KCl: 2 moles
  • O₂: 3 moles

The molar mass of the compounds is:

  • KClO₃: 122.45 g/mole
  • KCl: 74.45 g/mole
  • O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
  • KCl: 2 moles ×74.45 g/mole= 148.9 grams
  • O₂: 3 moles ×32 g/mole= 96 grams

<h3>Ideal gas law</h3>

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

  • P is the gas pressure.
  • V is the volume that the gas occupies.
  • T is the temperature of the gas.
  • R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
  • n is the number of moles of the gas.

<h3>Number of O₂ produced.</h3>

165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:

  • P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
  • V= 165 mL= 0.165 L (being 1000 mL= 1 L)
  • n= ?
  • R= 0.082 \frac{atmL}{molK}
  • T= 30 C= 303 K (being 0 C= 273 K)

Replacing in the ideal gas law:

0.888231 atm× 0.165 L = n× 0.082 \frac{atmL}{molK}× 303 K

Solving:

n= (0.888231 atm× 0.165 L)÷ (0.082 \frac{atmL}{molK}× 303 K)

<u><em>n= 0.0059 moles</em></u>

Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.

<h3>Mass of KClO₃ required</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?

mass of KClO_{3}= \frac{0.0059 moles of O_{2}x 244.8 grams of KClO_{3}}{3 moles of O_{2}}

<u><em>mass of KClO₃= 0.48144 grams</em></u>

Finally, 0.48144 grams of KClO₃ was decomposed.

Learn more about

the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

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