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Inessa05 [86]
3 years ago
7

Acid rain can be destructive to both the natural environment and human-made structures. The equation below shows a reaction that

occurs that may lead to the formation of acid rain. 3NO2 + H2O mc025-1.jpg 2HNO3 + NO How many moles (precise to the nearest 0.01 mol) of nitric acid are produced from 300.00 mol of nitrogen dioxide?
Chemistry
2 answers:
alexandr1967 [171]3 years ago
8 0
The moles of nitric acid formed can be computed using the stoichiometric relation obtained from the given equation which indicates that for every 3 moles of nitrogen dioxide (NO2), 2 moles of nitric acid (HNO3) is formed. 300 moles of NO2 is then multiplied to 2 mol HNO3/ 3 mol NO2. The answer is 200 moles of nitric acid is produced from 300 moles of NO2. 
VLD [36.1K]3 years ago
4 0

Answer;

200.00 moles of nitric acid is produced

Explanation;

From the equation;

3NO2 + H2O → 2HNO3 + NO

The mole ratio of nitrogen (IV) oxide  and nitric acid is 3 is to 2;

Thus; for every 3 moles of nitrogen dioxide (NO2), 2 moles of nitric acid (HNO3) are formed. 

Therefore; 300 moles of NO2 will produce;

2 mol HNO3/ 3 mol NO2 (300) = 200 moles

Hence, 200 moles of nitric acid are produced from 300 moles of NO2. 

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3 years ago
Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of
bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

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

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

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

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

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

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

<u>Step 4:</u> Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

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

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

<u>Step 6:</u> Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

<u>Step 7</u>: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

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

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0
3 years ago
An unknown compound contains 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by mass. A mass spectrometry analysis reveals that
tiny-mole [99]

Answer:

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The compound is 75.69 % C, 8.80 % H and 15.51 % O. This data means, that in 100 g of compound we have 75.69 g, 15.51 g and 8.80 g of, C, O and H, respectively. We know the molar mass of the compound, so we can work to solve the moles of each element.

In 100 g of compound we have 75.69 g C, 15.51 g O and 8.80 g H

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(412 . 8.80) / 100 = 36.2 g of H

Now, we can determine the moles of each, that are contained in 1 mol of compound.

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36 g / 1 g/mol = 36 H

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