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Nuetrik [128]
3 years ago
12

In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t

o allow the gas to reach room temperature. If the pressure gauge reads 173 psi and the lecture hall is 20°C, how many moles of sulfur dioxide are in the bottle? In order to solve this problem, you will first need to calculate the pressure of the gas. Hint: The gauge reads zero when 14.7 psi of gas remains.
Chemistry
1 answer:
Marrrta [24]3 years ago
6 0

Answer:

0.66 mol

Explanation:

Zero Gauge pressure = 14.7 psi

Pressure read = 173 psi

Actual pressure = 173 psi - 14.7 psi = 158.3 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 10.77 atm

Given that:

Temperature = 20 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (20+ 273.15) K = 298.15 K

V = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol  × 298.15 K

⇒n = 0.66 mol

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How many moles of NH3 would be formed from the complete reaction of 16.0 g H2?
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Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

N₂ + 3 H₂ → 2 NH₃

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:

  • N₂: 1 mole
  • H₂: 3 moles
  • NH₃: 2 moles

The molar mass of the compounds is:

  • N₂: 14 g/mole
  • H₂: 2 g/mole
  • NH₃: 17 g/mole

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

  • N₂: 1 mole ×14 g/mole= 14 grams
  • H₂: 3 moles ×2 g/mole= 6 grams
  • NH₃: 2 moles ×17 g/mole=34 grams

<h3>Mass of NH₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?

moles of NH_{3}= \frac{16 grams of H_{2} x2moles of NH_{3}}{6 grams of H_{2}}

<u><em>moles of NH₃= 5.33 moles</em></u>

Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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