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Juliette [100K]
3 years ago
9

2.0L of hydrogen gas is mixed with 3.0L of nitrogen gas at STP in a rigid 5.0L vessel. A reaction occurs, producing ammonia gas

NH3 and causing the pressure to change. How many moles of ammonia are produced? Hint: Write reaction and determine LR first How many moles of excess reactant remain in the vessel after the reaction? What is the final total pressure after the reaction if the temperature is kept at 273 K?
Chemistry
1 answer:
IrinaK [193]3 years ago
7 0

Answer:

Moles NH₃: 0.0593

0.104 moles of N₂ remain

Final pressure: 0.163atm

Explanation:

The reaction of nitrogen with hydrogen to produce ammonia is:

N₂ + 3 H₂ → 2 NH₃

Using PV = nRT, moles of N₂ and H₂ are:

N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂

H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂

The complete reaction of N₂ requires:

0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>

That means limiting reactant is H₂. And moles of NH₃ produced are:

0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>

Moles of N₂ remain are:

0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>

And final pressure is:

P = nRT / V

P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L

<em>P = 0.163atm</em>

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7 0
3 years ago
An isotope of carbon has 8 neutrons. What is the correct hyphenated symbol for this isotope?
Rom4ik [11]

Answer:

Carbon-14 or C-14

Explanation:

Since the isotope is a carbon isotope, it means the the atomic number of the isotope is 6.

The atomic number of an element is the proton number.

Therefore, the isotope has a proton number of 6

Now we need to obtain the mass number of isotope in order to write the symbol of the isotope. This is illustrated below:

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4 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

<u>Step 1:</u> Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

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andreyandreev [35.5K]
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Cylinder B - Adding 25 grams raised the water level by roughly 2.8 mL

Cylinder C - Adding 25 grams raised the water level by roughly 3.5 mL

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C. 25 g/3.5 mL = 7.1429 g/mL = Zinc

Hope I helped!

5 0
3 years ago
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